Free math worksheets

- Check each required function property separately rather than jumping to the improper integral. - For continuity, ask whether the denominator can be zero on the stated interval. - To verify decrease, determine the sign of the first derivative on the interval.

1. For \(x\ge1\), \(f(x)>0\). 2. The denominator never vanishes on \([1,\infty)\), so \(f\) is continuous there. 3. Since \(f'(x)=-\frac{2}{(x+2)^3}<0\) for \(x\ge1\), \(f\) is decreasing. 4. Therefore all three function conditions for the integral test are satisfied on \([1,\infty)\).

Yes. \(f\) is positive, continuous, and decreasing on \([1,\infty)\), so its function conditions for the integral test are satisfied.

- Choose a function \(f(x)\) whose values at positive integers equal the series terms. - Verify that \(f\) is positive, continuous, and decreasing on the required interval. - Evaluate the corresponding improper integral; the integral test classifies convergence but does not give the exact series sum.

1. Let \(f(x)=\frac1{x^2}\). Then \(f(n)=\frac1{n^2}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{2}{x^3}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}f(x)\,dx=\lim_{b\to\infty}\left[-\frac1x\right]_{1}^{b}=1\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=1/x^2\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\), and \(\int_1^{\infty}x^{-2}\,dx=1\). Therefore the series converges by the integral test.

- Replace the integer index with a real variable to define the associated function. - Verify positivity, continuity, and decrease on \([1,\infty)\); the shifted denominator does not change the power behavior. - Evaluate the improper integral using the antiderivative of \((x+3)^{-2}\).

1. Let \(f(x)=\frac1{(x+3)^2}\). Then \(f(n)=\frac1{(n+3)^2}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{2}{(x+3)^3}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}f(x)\,dx=\lim_{b\to\infty}\left[-\frac1{x+3}\right]_{1}^{b}=\frac14\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=1/(x+3)^2\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\), and \(\int_1^{\infty}(x+3)^{-2}\,dx=\frac14\). Therefore the series converges by the integral test.

- Use the shifted power function \(f(x)=(x+4)^{-3/2}\) and verify the integral-test hypotheses. - Apply the power rule to find an antiderivative. - Evaluate both endpoint contributions carefully to obtain the exact improper-integral value.

1. Let \(f(x)=\frac1{(x+4)^{3/2}}\). Then \(f(n)=\frac1{(n+4)^{3/2}}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{3}{2(x+4)^{5/2}}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}\frac1{(x+4)^{3/2}}\,dx=\lim_{b\to\infty}\left[-\frac2{\sqrt{x+4}}\right]_{1}^{b}=\frac2{\sqrt5}\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=1/(x+4)^{3/2}\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\). The improper integral equals \(\frac2{\sqrt5}\), so the series converges by the integral test.

- Define \(f(x)=\frac1{3x+2}\) and verify its sign, continuity, and derivative on \([1,\infty)\). - Integrate using the substitution \(u=3x+2\). - Determine whether the logarithmic antiderivative has a finite limit at infinity.

1. Let \(f(x)=\frac1{3x+2}\). Then \(f(n)=\frac1{3n+2}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac3{(3x+2)^2}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}\frac1{3x+2}\,dx=\lim_{b\to\infty}\left[\frac13\ln(3x+2)\right]_{1}^{b}=\infty\). 4. Therefore the series diverges by the integral test.

For \(f(x)=1/(3x+2)\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\), while \(\int_1^{\infty}\frac{dx}{3x+2}=\infty\). Therefore the series diverges by the integral test.

- Separate the conclusion “the improper integral is finite” from the numerical value of that integral. - Recall exactly what equivalence the integral test establishes between a series and an improper integral. - The test determines convergence behavior, not equality between the series sum and the integral value.

1. The associated function is positive, continuous, and decreasing on \([1,\infty)\), and the improper integral is finite. 2. Therefore the integral test proves that the series converges. 3. The integral test does not state that the series sum equals the improper integral, so the value \(1\) is not established as the series sum.

The series converges, but the integral test does not give its exact sum.

- Write the original series as a finite prefix plus the tail beginning at \(n=21\). - A finite sum cannot create or remove divergence in an infinite tail. - The same principle allows the integral test to begin at any later point where its hypotheses hold.

1. Split the full series into the finite sum \(\sum_{n=1}^{20}f(n)\) and the infinite tail \(\sum_{n=21}^{\infty}f(n)\). 2. The finite prefix has a finite value, so adding or removing it cannot change whether the tail converges. 3. Thus \(\sum_{n=1}^{\infty}f(n)\) and \(\sum_{n=21}^{\infty}f(n)\) have the same convergence behavior.

The two series have the same convergence behavior because they differ by only a finite sum.

- Confirm that the stated function satisfies every hypothesis of the integral test. - Use the finiteness of the improper integral to classify the series tail. - Distinguish the integral’s numerical value from the unknown sum of the discrete series.

1. All integral-test hypotheses hold on \([4,\infty)\). 2. Because the improper integral is finite, \(\sum_{n=4}^{\infty}a_n\) converges. 3. The integral test does not imply that the series sum equals \(\frac73\).

The integral test proves convergence, but it does not determine the series sum as \(\frac73\).

- Differentiate \(f\) to determine whether it is decreasing on the stated interval. - The standard integral test requires a positive, continuous, decreasing associated function. - Independently check the term limit to see the actual behavior of the series.

1. \(f'(x)=\frac1{x^2}>0\), so \(f\) is increasing rather than decreasing. 2. Therefore a required integral-test hypothesis fails. 3. In fact, \(a_n\to2\ne0\), so the series diverges by the \(n\)th-term test.

The function is increasing, not decreasing, so the integral test does not apply. In fact, the series diverges because \(a_n\to2\ne0\).

- Write the natural real-variable extension of the summand. - Check the sign of the sine factor over the entire interval, not only at a few integers. - Decide whether the oscillating function can be decreasing on all sufficiently large inputs.

1. The natural associated function is \(f(x)=\frac{\sin x}{x^2}\), which is continuous for \(x\ge1\). 2. Because \(\sin x\) changes sign, \(f\) is not positive on the entire interval. 3. The oscillation also prevents \(f\) from being decreasing on the entire interval. 4. Therefore the standard integral test does not apply to this associated function.

Positivity and monotonic decrease both fail, so the standard integral test does not apply.

- Evaluate the logarithm in the denominator at the proposed starting point. - A valid integral-test function must be defined and continuous on the full interval of integration. - Check that beginning at \(2\) avoids the domain problem and gives a valid tail.

1. \(\ln1=0\), so the term \(\frac1{1\ln1}\) is undefined. 2. Likewise, the associated function \(f(x)=\frac1{x\ln x}\) is not defined or continuous at \(x=1\). 3. On \([2,\infty)\), the function is positive, continuous, and decreasing, so the integral-test setup is valid there.

The term and associated function are undefined at \(1\); the setup must begin at \(2\) or later.

- Separate each series into a finite initial part and an infinite tail. - Identify where the two tails become identical. - Convergence is unaffected by changing finitely many terms.

1. The integral test shows that the tail of the first series beginning after index \(50\) converges. 2. The two series have identical terms from index \(51\) onward, so the second series has the same convergent tail. 3. The first \(50\) terms of the second series form a finite sum. 4. Adding finitely many terms to a convergent tail preserves convergence.

The second series converges because it has the same convergent tail as the first series.

- Look at which upper corner of each rectangle touches the curve. - After identifying the endpoint rule, use the corresponding x-values to determine the rectangle heights. - Compare the rectangle tops with the curve across each subinterval before choosing the inequality direction.

1. In each rectangle, the upper-right corner lies on the graph of \(f\), so the rectangle heights are set by the right endpoints \(x=2,3,4,5\). 2. Each subinterval has width \(1\), so the total rectangle area is \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}=\sum_{n=2}^{5}\frac{1}{n^2}\). 3. The graph shows that \(f\) is decreasing and that these right-endpoint rectangles lie below the curve on their subintervals. 4. Therefore \(\sum_{n=2}^{5}\frac{1}{n^2}<\int_1^5\frac{1}{x^2}\,dx\).

a) Right endpoints. b) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}=\sum_{n=2}^{5}\frac{1}{n^2}\). c) \(\sum_{n=2}^{5}\frac{1}{n^2}<\int_1^5\frac{1}{x^2}\,dx\).

- For monotonicity, differentiate \(f(x)=\frac1{x\ln x}\) or show its positive denominator increases for \(x\ge2\). - In the improper integral, use the substitution \(u=\ln x\), so \(du=\frac{dx}{x}\). - Determine whether the resulting logarithmic antiderivative stays finite as the upper limit grows.

1. Let \(f(x)=\frac1{x\ln x}\). Then \(f(n)=\frac1{n\ln n}\), and \(f\) is positive and continuous for \(x\ge2\). 2. Since \(f'(x)=-\frac{\ln x+1}{x^2(\ln x)^2}<0\) for \(x\ge2\), \(f\) is decreasing there. 3. \(\int_{2}^{\infty}\frac1{x\ln x}\,dx=\lim_{b\to\infty}\left[\ln(\ln x)\right]_{2}^{b}=\infty\). 4. Therefore the series diverges by the integral test.

For \(f(x)=1/(x\ln x)\), \(f\) is positive, continuous, and decreasing on \([2,\infty)\), while \(\int_2^{\infty}\frac{dx}{x\ln x}=\infty\). Therefore the series diverges by the integral test.

- Check that \(f(x)=\frac1{x(\ln x)^2}\) is positive, continuous, and decreasing for \(x\ge2\). - Use \(u=\ln x\) to transform the improper integral into a power integral in \(u\). - Decide convergence by whether that transformed integral has a finite value at infinity.

1. Let \(f(x)=\frac1{x(\ln x)^2}\). Then \(f(n)=\frac1{n(\ln n)^2}\), and \(f\) is positive and continuous for \(x\ge2\). 2. Since \(f'(x)=-\frac{\ln x+2}{x^2(\ln x)^3}<0\) for \(x\ge2\), \(f\) is decreasing there. 3. \(\int_{2}^{\infty}\frac1{x(\ln x)^2}\,dx=\lim_{b\to\infty}\left[-\frac1{\ln x}\right]_{2}^{b}=\frac1{\ln2}\), which is finite. 4. Therefore the integral test applies and the series converges.

Yes. For \(f(x)=1/[x(\ln x)^2]\), \(f\) is positive, continuous, and decreasing on \([2,\infty)\), and \(\int_2^{\infty}\frac{dx}{x(\ln x)^2}=\frac1{\ln2}\). Therefore the series converges.

- Verify that \(f(x)=\frac1{x\sqrt{\ln x}}\) satisfies the integral-test hypotheses for \(x\ge2\). - Substitute \(u=\ln x\) in the improper integral. - Examine the growth of the resulting antiderivative as the upper limit tends to infinity.

1. Let \(f(x)=\frac1{x\sqrt{\ln x}}\). Then \(f(n)=\frac1{n\sqrt{\ln n}}\), and \(f\) is positive and continuous for \(x\ge2\). 2. Since \(f'(x)=-\frac{\ln x+\frac12}{x^2(\ln x)^{3/2}}<0\) for \(x\ge2\), \(f\) is decreasing there. 3. \(\int_{2}^{\infty}\frac1{x\sqrt{\ln x}}\,dx=\lim_{b\to\infty}\left[2\sqrt{\ln x}\right]_{2}^{b}=\infty\). 4. Therefore the series diverges by the integral test.

For \(f(x)=1/[x\sqrt{\ln x}]\), \(f\) is positive, continuous, and decreasing on \([2,\infty)\), while \(\int_2^{\infty}\frac{dx}{x\sqrt{\ln x}}=\infty\). Therefore the series diverges by the integral test.

- Use \(f(x)=\frac{x}{(x^2+1)^2}\) so that \(f(n)\) matches the summand. - Differentiate to verify that \(f\) is decreasing on the interval used by the series. - For the improper integral, let \(u=x^2+1\) and determine whether the resulting value is finite.

1. Let \(f(x)=\frac{x}{(x^2+1)^2}\). Then \(f(n)=\frac{n}{(n^2+1)^2}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=\frac{1-3x^2}{(x^2+1)^3}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}\frac{x}{(x^2+1)^2}\,dx=\lim_{b\to\infty}\left[-\frac1{2(x^2+1)}\right]_{1}^{b}=\frac14\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=x/(x^2+1)^2\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\), and \(\int_1^{\infty}\frac{x}{(x^2+1)^2}\,dx=\frac14\). Therefore the series converges by the integral test.

- Use \(f(x)=\frac{x}{x^2+1}\) and verify that it is positive, continuous, and decreasing for \(x\ge1\). - In the improper integral, let \(u=x^2+1\). - Examine the logarithmic antiderivative as the upper limit tends to infinity.

1. Let \(f(x)=\frac{x}{x^2+1}\). Then \(f(n)=\frac{n}{n^2+1}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=\frac{1-x^2}{(x^2+1)^2}\le0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}\frac{x}{x^2+1}\,dx=\lim_{b\to\infty}\left[\frac12\ln(x^2+1)\right]_{1}^{b}=\infty\). 4. Therefore the series diverges by the integral test.

For \(f(x)=x/(x^2+1)\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\), while \(\int_1^{\infty}\frac{x}{x^2+1}\,dx=\infty\). Therefore the series diverges by the integral test.

- Verify the integral-test hypotheses for \(f(x)=\frac1{x^2+4}\) on \([1,\infty)\). - Match the denominator to the standard antiderivative for \(\frac1{x^2+a^2}\). - Determine whether the arctangent expression has a finite limit at infinity.

1. Let \(f(x)=\frac1{x^2+4}\). Then \(f(n)=\frac1{n^2+4}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{2x}{(x^2+4)^2}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}\frac1{x^2+4}\,\mathrm{d}x=\lim_{b\to\infty}\left[\frac12\arctan\left(\frac{x}{2}\right)\right]_{1}^{b}=\frac{\pi}{4}-\frac12\arctan\left(\frac12\right)\), which is finite. 4. Therefore the integral test applies and the series converges.

Yes. For \(f(x)=\frac1{x^2+4}\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\), and the improper integral equals \(\frac{\pi}{4}-\frac12\arctan\left(\frac12\right)\). Therefore the series converges.

- Check positivity, continuity, and decrease for \(f(x)=\frac1{\sqrt{x^2+1}}\). - Use an antiderivative such as \(\ln\left(x+\sqrt{x^2+1}\right)\). - Decide whether that expression remains bounded as the upper integration limit grows.

1. Let \(f(x)=\frac1{\sqrt{x^2+1}}\). Then \(f(n)=\frac1{\sqrt{n^2+1}}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{x}{(x^2+1)^{3/2}}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}\frac1{\sqrt{x^2+1}}\,dx=\lim_{b\to\infty}\left[\ln\left(x+\sqrt{x^2+1}\right)\right]_{1}^{b}=\infty\). 4. Therefore the series diverges by the integral test.

For \(f(x)=1/\sqrt{x^2+1}\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\), while the associated improper integral diverges. Therefore the series diverges by the integral test.

- Use \(f(x)=\frac1{x(x+1)}\) and verify positivity, continuity, and decrease on \([1,\infty)\). - Decompose the integrand into partial fractions before integrating. - Invoke the integral test only after the hypotheses and finiteness of the improper integral are established.

1. Let \(f(x)=\frac1{x(x+1)}\). Then \(f(n)=\frac1{n(n+1)}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{2x+1}{(x^2+x)^2}<0\) for \(x\ge1\), \(f\) is decreasing, so the integral test applies. 3. \(\int_{1}^{\infty}\frac1{x(x+1)}\,dx=\lim_{b\to\infty}\left[\ln x-\ln(x+1)\right]_{1}^{b}=\ln2\), which is finite. 4. Therefore the series converges by the integral test.

The associated function is positive, continuous, and decreasing on \([1,\infty)\), so the integral test applies. Since the improper integral equals \(\ln2\), the series converges.

- Check the integral-test hypotheses for \(f(x)=\frac1{x^2-1}\) on \([2,\infty)\), where the denominator is nonzero. - Factor \(x^2-1\) and use partial fractions to integrate. - Evaluate the logarithmic endpoint limit carefully to obtain the exact improper-integral value.

1. Let \(f(x)=\frac1{x^2-1}\). Then \(f(n)=\frac1{n^2-1}\), and \(f\) is positive and continuous for \(x\ge2\). 2. Since \(f'(x)=-\frac{2x}{(x^2-1)^2}<0\) for \(x\ge2\), \(f\) is decreasing there. 3. \(\int_{2}^{\infty}\frac1{x^2-1}\,dx=\lim_{b\to\infty}\left[\frac12\ln\left(\frac{x-1}{x+1}\right)\right]_{2}^{b}=\frac12\ln3\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=1/(x^2-1)\), \(f\) is positive, continuous, and decreasing on \([2,\infty)\). The improper integral equals \(\frac12\ln3\), so the series converges by the integral test.

- Define the matching real-variable function and verify that it is positive, continuous, and decreasing for \(x\ge1\). - Complete the square: \(x^2+2x+2=(x+1)^2+1\). - Use the arctangent antiderivative and check whether its value at infinity is finite.

1. Let \(f(x)=\frac1{x^2+2x+2}\). Then \(f(n)=\frac1{n^2+2n+2}\), and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{2x+2}{(x^2+2x+2)^2}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. \(\int_{1}^{\infty}\frac1{x^2+2x+2}\,dx=\lim_{b\to\infty}\left[\arctan(x+1)\right]_{1}^{b}=\frac{\pi}{2}-\arctan2\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=1/(x^2+2x+2)\), \(f\) is positive, continuous, and decreasing on \([1,\infty)\), and the improper integral equals \(\frac{\pi}{2}-\arctan2\). Therefore the series converges.

- Verify the integral-test hypotheses for \(f(x)=\frac1{x(\ln x)^3}\) on \([2,\infty)\). - Use the substitution \(u=\ln x\), so \(du=\frac{dx}{x}\). - Evaluate the resulting negative-power integral at infinity and decide whether it is finite.

1. Let \(f(x)=\frac1{x(\ln x)^3}\). Then \(f(n)=\frac1{n(\ln n)^3}\), and \(f\) is positive and continuous for \(x\ge2\). 2. Since \(f'(x)=-\frac{\ln x+3}{x^2(\ln x)^4}<0\) for \(x\ge2\), \(f\) is decreasing there. 3. \(\int_{2}^{\infty}\frac1{x(\ln x)^3}\,dx=\lim_{b\to\infty}\left[-\frac1{2(\ln x)^2}\right]_{2}^{b}=\frac1{2(\ln2)^2}\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=1/[x(\ln x)^3]\), \(f\) is positive, continuous, and decreasing on \([2,\infty)\), and the improper integral equals \(\frac1{2(\ln2)^2}\). Therefore the series converges.

- For \(f(x)=\frac{\ln x}{x^2}\), verify positivity and continuity, then use the derivative to check decrease for \(x\ge2\). - Evaluate the improper integral by integration by parts, taking \(u=\ln x\). - Determine whether the boundary term at infinity vanishes before classifying the series.

1. Let \(f(x)=\frac{\ln x}{x^2}\). Then \(f(n)=\frac{\ln n}{n^2}\), and \(f\) is positive and continuous for \(x\ge2\). 2. Since \(f'(x)=\frac{1-2\ln x}{x^3}<0\) for \(x\ge2\), \(f\) is decreasing there. 3. \(\int_{2}^{\infty}\frac{\ln x}{x^2}\,dx=\lim_{b\to\infty}\left[-\frac{\ln x+1}{x}\right]_{2}^{b}=\frac{\ln2+1}{2}\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=\ln x/x^2\), \(f\) is positive, continuous, and decreasing on \([2,\infty)\), and the improper integral equals \(\frac{\ln2+1}{2}\). Therefore the series converges.

- Verify the integral-test hypotheses for \(f(x)=\frac1{x(\ln x+1)}\) beginning at \(x=1\). - Use the substitution \(u=\ln x+1\), so \(du=\frac{dx}{x}\). - Determine whether the resulting logarithmic integral remains finite as its upper limit grows.

1. Let \(f(x)=\frac1{x(\ln x+1)}\). Then \(f(n)\) equals the summand, and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{\ln x+2}{x^2(\ln x+1)^2}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. With \(u=\ln x+1\), \(\int_{1}^{\infty}\frac1{x(\ln x+1)}\,dx=\int_{1}^{\infty}\frac1u\,du=\infty\). 4. Therefore the series diverges by the integral test.

The associated function is positive, continuous, and decreasing on \([1,\infty)\), and the improper integral is infinite. Therefore the series diverges by the integral test.

- Check the value of \(\ln x+1\) at and beyond \(x=1\) to justify positivity and continuity. - Differentiate the associated function to verify that it is decreasing on the full interval. - Use \(u=\ln x+1\) to convert the improper integral to a power integral.

1. Let \(f(x)=\frac1{x(\ln x+1)^2}\). For \(x\ge1\), \(\ln x+1\ge1\), so \(f\) is positive and continuous and the starting point is valid. 2. Since \(f'(x)=-\frac{\ln x+3}{x^2(\ln x+1)^3}<0\) for \(x\ge1\), \(f\) is decreasing there. 3. With \(u=\ln x+1\), \(\int_{1}^{\infty}\frac1{x(\ln x+1)^2}\,dx=\int_{1}^{\infty}u^{-2}\,du=1\), which is finite. 4. Therefore the series converges by the integral test.

The starting point \(x=1\) is valid because \(\ln x+1\ge1\) there and beyond. The associated function is positive, continuous, and decreasing, the improper integral equals \(1\), and the series converges.

- Differentiate \(f(x)=\frac{x}{(x^2+9)^2}\) to find an integer index beyond which it decreases. - Apply the integral test to that tail; finitely many earlier terms do not affect convergence. - Use \(u=x^2+9\) to evaluate the tail’s improper integral.

1. Let \(f(x)=\frac{x}{(x^2+9)^2}\). Then \(f(n)\) equals the summand, and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=\frac{3(3-x^2)}{(x^2+9)^3}<0\) for \(x\ge2\), the tail beginning at \(n=2\) is decreasing. 3. \(\int_{2}^{\infty}\frac{x}{(x^2+9)^2}\,dx=\frac1{26}\), which is finite. 4. The tail converges by the integral test, and adding the first term does not change convergence. Therefore the original series converges.

The associated function is decreasing for \(x\ge2\), and the tail integral equals \(\frac1{26}\). Thus the tail, and therefore the original series, converges.

- Differentiate \(f(x)=\frac{x}{x^2+9}\) to locate a tail on which it is decreasing. - The integral test may be applied after discarding finitely many initial terms. - Use \(u=x^2+9\) and examine the logarithmic antiderivative at infinity.

1. Let \(f(x)=\frac{x}{x^2+9}\). Then \(f(n)\) equals the summand, and \(f\) is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=\frac{9-x^2}{(x^2+9)^2}\le0\) for \(x\ge3\), the tail beginning at \(n=3\) is decreasing. 3. \(\int_{3}^{\infty}\frac{x}{x^2+9}\,dx=\lim_{b\to\infty}\left[\frac12\ln(x^2+9)\right]_{3}^{b}=\infty\). 4. The tail diverges by the integral test, so the original series diverges.

The associated function is positive and continuous for \(x\ge1\) and decreasing for \(x\ge3\). The tail integral is infinite, so the tail and the original series diverge by the integral test.

- Positivity and continuity hold for positive \(x\); the limiting condition is monotonicity. - Differentiate the quotient and determine where the derivative is nonpositive. - Choose the least integer whose entire right-hand interval lies in that region.

1. \(f'(x)=\frac{9-x^2}{(x^2+9)^2}\). 2. The derivative is nonpositive when \(x\ge3\). 3. The function is positive and continuous for \(x>0\), so the least integer is \(N=3\).

\(N=3\).

- Locate every point where the proposed associated function is undefined. - Match the interval for the integral test to the actual starting index of the series. - After repairing the interval, evaluate the shifted reciprocal-square integral.

1. The function is not continuous at \(x=2\), so it cannot be used on \([1,\infty)\). 2. The series begins at \(n=3\), and \(f\) is positive, continuous, and decreasing on \([3,\infty)\). 3. \(\int_3^\infty\frac1{(x-2)^2}\,\mathrm{d}x=1\), which is finite, so the series converges by the integral test.

The proposed interval crosses the discontinuity at \(x=2\). Start at \(x=3\); the improper integral equals \(1\), so the series converges.

- Matching the sequence at integer inputs is necessary but not sufficient for an integral-test extension. - Examine how the sine term behaves between consecutive integers. - Check positivity and monotonicity on the entire interval, not only at integer points.

1. Although \(g(n)=a_n\) at every positive integer, agreement at integer inputs is not the only integral-test requirement. 2. The sine term makes \(g\) oscillate; for large \(x\), \(g\) is sometimes negative and it is not decreasing on all of \([1,\infty)\). 3. Therefore the integral test cannot be applied directly to \(g\). The standard extension \(f(x)=\frac1{x^2}\) satisfies the hypotheses.

No. The proposed extension is not positive and decreasing on \([1,\infty)\), so the integral-test hypotheses fail.

- Treat \(c\) as a constant when evaluating the shifted reciprocal-square integral. - Set the resulting expression equal to the given integral value and solve for \(c\). - After finding \(c\), verify the sign and monotonicity needed for the integral test.

1. \(\int_1^\infty\frac{c}{(x+1)^2}\,dx=\frac{c}{2}\). 2. Set \(\frac{c}{2}=3\), so \(c=6\). 3. For \(c=6\), the associated function is positive, continuous, and decreasing on \([1,\infty)\), and its improper integral is finite. 4. Therefore the series converges by the integral test.

\(c=6\), and the series converges.

- How can the square root in the exponent be replaced by a simpler variable? - Track how the differential combines with the remaining radical factor. - Does the resulting tail have finite total area?

1. Let \(f(x)=\frac{e^{-\sqrt x}}{\sqrt x}\). It is positive and continuous for \(x\ge1\), and both factors make it decrease as \(x\) grows. 2. Set \(u=\sqrt x\), so \(x=u^2\) and \(dx=2u\,du\). 3. Then \(\int_1^{\infty}\frac{e^{-\sqrt x}}{\sqrt x}\,dx=2\int_1^{\infty}e^{-u}\,du=\frac{2}{e}\). 4. The integral is finite, so the series converges.

The series converges. The associated improper integral equals \(\frac{2}{e}\).

- Which repeated radical can serve as a simpler variable? - How does the differential interact with the radical already in the denominator? - Does the simplified accumulated area remain bounded?

1. Let \(f(x)=\frac{1}{\sqrt x(1+\sqrt x)}\). It is positive, continuous, and decreasing for \(x\ge1\). 2. Set \(u=\sqrt x\), so \(dx=2u\,du\). 3. The improper integral becomes \(\int_1^{\infty}f(x)\,dx=2\int_1^{\infty}\frac{1}{1+u}\,du\), which diverges. 4. Therefore the series diverges by the integral test.

The series diverges.

- Compare the areas under a decreasing curve with left- and right-endpoint rectangles. - Use both the integral value and the first discrete term. - Check whether the claimed total lies inside the resulting interval.

1. For a positive decreasing function, \(\int_1^{\infty}f(x)\,\mathrm{d}x\le\sum_{n=1}^{\infty}f(n)\le f(1)+\int_1^{\infty}f(x)\,\mathrm{d}x\). 2. Substituting the given values gives \(5\le\sum_{n=1}^{\infty}f(n)\le1+5=6\). 3. The claimed sum \(7\) lies outside this required interval.

The claim cannot be true. The series sum must lie in \([5,6]\).

- Does the full denominator keep increasing on the stated interval? - Which inner quantity combines naturally with the reciprocal input factor? - What finite or infinite endpoint behavior remains after simplification?

1. Let \(f(x)=\frac{1}{x[1+(\ln x)^2]}\). It is positive and continuous for \(x\ge1\). 2. Its derivative is \(f'(x)=-\frac{(\ln x+1)^2}{x^2[1+(\ln x)^2]^2}\le0\), so \(f\) is decreasing. 3. With \(u=\ln x\) and \(du=\frac{dx}{x}\), \(\int_1^{\infty}f(x)\,dx=\int_0^{\infty}\frac{1}{1+u^2}\,du=\frac{\pi}{2}\). 4. The integral is finite, so the series converges by the integral test.

The series converges. The associated improper integral equals \(\frac{\pi}{2}\).

- Identify which repeated fractional power can serve as a simpler variable. - Track how the differential interacts with the power already in the denominator. - Check whether the transformed tail has a finite accumulated value.

1. Let \(f(x)=\frac{1}{x^{2/3}(1+x^{1/3})^2}\). It is positive and continuous for \(x\ge1\). 2. Both positive factors in the denominator increase on \([1,\infty)\), so \(f\) is decreasing. 3. Set \(u=x^{1/3}\), so \(x=u^3\) and \(dx=3u^2\,du\). Then \(\int_1^{\infty}f(x)\,dx=3\int_1^{\infty}\frac{1}{(1+u)^2}\,du=\frac32\). 4. The integral is finite, so the series converges by the integral test.

The series converges. The associated improper integral equals \(\frac32\).

- Which expression in the denominator changes at a rate matching the numerator? - Rewrite the accumulated area using that relationship. - Evaluate only the endpoint behavior needed for convergence.

1. Let \(f(x)=\frac{1}{e^x+1}\). It is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{e^x}{(e^x+1)^2}<0\), the function is decreasing. 3. Rewrite \(f(x)=\frac{e^{-x}}{1+e^{-x}}\). With \(u=e^{-x}\), \(\int_1^{\infty}f(x)\,dx=\int_0^{e^{-1}}\frac{1}{1+u}\,du=\ln(1+e^{-1})\). 4. The integral is finite, so the series converges by the integral test.

The series converges. The associated improper integral equals \(\ln(1+e^{-1})\).

- List every condition required by the stated result. - Identify which condition Min-jun did not establish. - Distinguish a flaw in the argument from the actual behavior of the series.

1. The integral test requires a positive, continuous function that is decreasing on a tail and agrees with the sequence at integers. 2. Positivity and continuity alone are insufficient. 3. The derivative is \(f'(x)=\frac{x\cos x-2(2+\sin x)}{x^3}\). 4. At \(x=2\pi k\), where \(k\) is any positive integer, \(f'(x)=\frac{2\pi k-4}{(2\pi k)^3}>0\). 5. These points occur arbitrarily far out, so \(f\) is not decreasing on any tail. Therefore this extension does not satisfy the integral-test hypotheses, and Min-jun's argument is invalid.

The argument is incomplete because eventual decrease was not established. In fact, the chosen extension is not decreasing on any tail, so the integral test cannot be applied to it.

- Keep the upper and lower endpoint values separate until the final subtraction. - What value does the inverse-tangent term approach at the infinite endpoint? - Check the sign of each contribution before combining them.

1. The continuous extension \(f(x)=1/(x^4+x^2)\) is positive, continuous, and decreasing for \(x\ge1\). 2. An antiderivative is \(-\frac1x-\arctan x\). 3. At infinity, the antiderivative approaches \(-\pi/2\); at \(x=1\), it equals \(-1-\pi/4\). 4. Subtracting lower from upper gives \(\int_1^{\infty}\frac{1}{x^4+x^2}\,\mathrm{d}x=1-\frac{\pi}{4}\), not \(1+\frac{\pi}{4}\). 5. The integral is finite, so the series converges.

Mariam used the wrong sign when subtracting endpoint values. The correct integral is \(1-\frac{\pi}{4}\), and the series converges.

- Which denominator expression has a change rate represented in the numerator? - Use the parameter restrictions when interpreting the transformed endpoints. - What equation connects the prescribed area with the unknown parameter?

1. For \(c>0\), let \(f_c(x)=\frac{c}{x^2+c^2}\). It is positive, continuous, and decreasing for \(x\ge1\). 2. The associated integral is \(\int_1^{\infty}\frac{c}{x^2+c^2}\,dx=\left[\arctan\!\left(\frac{x}{c}\right)\right]_1^{\infty}=\arctan c\). 3. Solving \(\arctan c=\frac{\pi}{3}\) gives \(c=\sqrt3\). 4. The prescribed integral is finite, so the resulting series converges by the integral test.

\(c=\sqrt3\), and the resulting series converges.

- Check the domain carefully before choosing the starting point. - Look for an inverse-trigonometric expression involving the reciprocal of the variable. - Evaluate the reciprocal inside that expression at the infinite endpoint.

1. Let \(f(x)=\frac{1}{x\sqrt{x^2-1}}\). It is positive and continuous for \(x\ge2\). 2. Its denominator is increasing on \([2,\infty)\), so \(f\) is decreasing. 3. Since \(\frac{d}{dx}\arccos\!\left(\frac1x\right)=\frac{1}{x\sqrt{x^2-1}}\), \(\int_2^{\infty}f(x)\,dx=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}\). 4. The integral is finite, so the series converges by the integral test.

The series converges. The associated improper integral equals \(\frac{\pi}{6}\).

- Verify that the continuous extension is decreasing on the required interval. - Use the inequality \(\arctan y\le y\) for nonnegative \(y\). - Compare the resulting improper integral with the integral of \(1/x^2\).

1. Let \(f(x)=\arctan\!\left(\frac{1}{x^2}\right)\). It is positive and continuous for \(x\ge1\). 2. Its derivative is \(f'(x)=-\frac{2x}{x^4+1}<0\), so \(f\) is decreasing. 3. For \(y\ge0\), \(\arctan y\le y\). Therefore \(0\le f(x)\le\frac{1}{x^2}\). 4. Since \(\int_1^{\infty}x^{-2}\,dx\) is finite, the associated integral for \(f\) converges. Thus the series converges by the integral test.

The series converges.

- Differentiate the continuous extension to verify that it decreases. - Find an antiderivative by separating \(\ln(x+1)-\ln x\). - Examine the antiderivative at the infinite endpoint.

1. Let \(f(x)=\ln\!\left(1+\frac1x\right)\). It is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{1}{x(x+1)}<0\), the function is decreasing. 3. An antiderivative is \((x+1)\ln(x+1)-x\ln x\). 4. As \(x\to\infty\), this antiderivative grows without bound, so \(\int_1^{\infty}f(x)\,dx\) diverges. Therefore the series diverges by the integral test.

The series diverges.

- Use \(u=\sqrt{\ln x}\) to simplify both logarithmic terms. - Track how \(dx/x\) changes under the substitution. - Determine the behavior of the resulting logarithmic improper integral.

1. Let \(f(x)=\frac{1}{x(\ln x+\sqrt{\ln x})}\). It is positive and continuous for \(x\ge2\), and its denominator is increasing, so \(f\) is decreasing. 2. Set \(u=\sqrt{\ln x}\). Then \(\ln x=u^2\) and \(\frac{dx}{x}=2u\,du\). 3. The improper integral becomes \(\int_2^{\infty}f(x)\,dx=2\int_{\sqrt{\ln2}}^{\infty}\frac{1}{u+1}\,du\), which diverges. 4. Therefore the series diverges by the integral test.

The series diverges.

- Which logarithmic quantity simplifies the reciprocal input factor? - Compare the transformed denominator with its leading term for large inputs. - Does the accumulated tail remain bounded or grow without limit?

1. Let \(f(x)=\frac{1}{x\sqrt{1+(\ln x)^2}}\). It is positive and continuous for \(x\ge1\). 2. The denominator is increasing on \([1,\infty)\), so \(f\) is decreasing. 3. With \(u=\ln x\), \(\int_1^{\infty}f(x)\,dx=\int_0^{\infty}\frac{du}{\sqrt{1+u^2}}\). 4. An antiderivative is \(\ln(u+\sqrt{1+u^2})\), which is unbounded. Thus the integral and the series diverge.

The series diverges.

- Use the area under the matching power function to bound the omitted tail. - Translate the tolerance into an inequality for the stopping index. - The requested inequality is strict.

1. For \(f(x)=1/x^2\), the integral remainder estimate gives \(R_N\le\int_N^{\infty}x^{-2}\,dx=1/N\). 2. To guarantee \(R_N<0.005\), require \(1/N<0.005=1/200\). 3. Thus \(N>200\). 4. The least integer satisfying the estimate is \(N=201\).

The least integer guaranteed by the estimate is \(N=201\).

- Check how the continuous extension changes as its input grows. - Evaluate the accumulated area rather than comparing only the integrand. - Use the rectangle-area inequality associated with a decreasing function.

1. Let \(f(x)=1/\sqrt{x^2+4}\). It is positive, continuous, and decreasing for \(x\ge1\). 2. An antiderivative is \(\ln(x+\sqrt{x^2+4})\). 3. The improper integral diverges because \(\ln(x+\sqrt{x^2+4})\to\infty\). 4. Since \(f\) is decreasing, \(\sum_{n=1}^{N}f(n)\ge\int_1^{N+1}f(x)\,dx\), so \(S_N\ge\ln(N+1+\sqrt{(N+1)^2+4})-\ln(1+\sqrt5)\).

The series diverges, and \(S_N\ge\ln(N+1+\sqrt{(N+1)^2+4})-\ln(1+\sqrt5)\).

- Look at the series term as a function of a real variable and ask whether its tail has the standard properties needed for a function-based convergence test. - Check positivity, continuity, and eventual decrease before evaluating any associated improper integral. - The nested factor \(\ln x\) suggests a substitution that removes the factor \(1/x\).

1. Let \(f(x)=\frac{1}{x(\ln x)^2}\). For \(x\ge2\), \(f\) is positive and continuous. 2. Since \(x(\ln x)^2\) is increasing for \(x\ge2\), \(f\) is decreasing there. 3. The associated improper integral is \(\int_2^{\infty}\frac{dx}{x(\ln x)^2}\). With \(u=\ln x\), it becomes \(\int_{\ln2}^{\infty}u^{-2}\,du\), which converges. 4. Therefore the series converges by the integral test.

The series converges by the integral test.

- Define a real-variable function that agrees with the summand at integer inputs and verify positivity and continuity on \([3,\infty)\). - Differentiate the function and use \(\ln x>1\) for \(x\ge3\) to establish decrease. - Evaluate the improper integral by integration by parts twice; its finiteness determines the series behavior.

1. Let \(f(x)=\frac{(\ln x)^2}{x^2}\). Then \(f(n)\) equals the summand, and \(f\) is positive and continuous for \(x\ge3\). 2. Since \(f'(x)=\frac{2\ln x(1-\ln x)}{x^3}<0\) for \(x\ge3\), \(f\) is decreasing there. 3. Integrating by parts twice gives \(\int_{3}^{\infty}\frac{(\ln x)^2}{x^2}\,dx=\frac{(\ln3)^2+2\ln3+2}{3}\), which is finite. 4. Therefore the series converges by the integral test.

For \(f(x)=(\ln x)^2/x^2\), \(f\) is positive, continuous, and decreasing on \([3,\infty)\). The improper integral equals \(\frac{(\ln3)^2+2\ln3+2}{3}\), so the series converges.

- Differentiate the associated function to verify that it becomes decreasing for every fixed real \(q\). - Use \(u=\ln x\) to transform the improper integral into \(\int u^{-q}\,du\). - Apply the power-integral criterion at infinity to determine the parameter range.

1. Let \(f(x)=\frac1{x(\ln x)^q}\). For \(x\ge2\), \(f\) is positive and continuous. 2. Since \(f'(x)=-\frac{\ln x+q}{x^2(\ln x)^{q+1}}\), the function is decreasing once \(x\) is large enough that \(\ln x+q>0\). 3. With \(u=\ln x\), \(\int_{2}^{\infty}\frac1{x(\ln x)^q}\,dx=\int_{\ln2}^{\infty}u^{-q}\,du\), which converges exactly when \(q>1\). 4. Therefore the series converges for \(q>1\) and diverges for \(q\le1\).

The series converges for \(q>1\) and diverges for \(q\le1\).

- Differentiate the associated function; its eventual monotonicity depends on whether \(p\) is greater than \(\frac12\). - When the integral test applies, use \(u=x^2+1\) and analyze the resulting power integral. - For the remaining parameter values, check whether the original terms even approach \(0\).

1. Let \(f(x)=\frac{x}{(x^2+1)^p}\). It is positive and continuous for \(x\ge1\), and \(f'(x)=\frac{1+(1-2p)x^2}{(x^2+1)^{p+1}}\). 2. If \(p>\frac12\), then \(f\) is eventually decreasing. With \(u=x^2+1\), \(\int_1^\infty\frac{x}{(x^2+1)^p}\,dx=\frac12\int_2^\infty u^{-p}\,du\), which converges exactly when \(p>1\). 3. If \(0<p<\frac12\), then \(\frac{n}{(n^2+1)^p}=\frac{n^{1-2p}}{(1+1/n^2)^p}\to\infty\); if \(p=\frac12\), the terms approach \(1\). These cases diverge by the \(n\)th-term test. 4. Therefore the series converges for \(p>1\) and diverges for \(0<p\le1\).

The series converges for \(p>1\) and diverges for \(0<p\le1\).

- For \(p>0\), verify that the shifted power function satisfies all integral-test hypotheses. - Translate the integral by setting \(u=x+c\), then apply the power-integral criterion. - For \(p\le0\), examine the term limit before attempting to use the integral test.

1. If \(p>0\), let \(f(x)=(x+c)^{-p}\). It is positive and continuous for \(x\ge1\), and \(f'(x)=-p(x+c)^{-p-1}<0\), so it is decreasing. 2. The improper integral \(\int_1^\infty(x+c)^{-p}\,dx=\int_{1+c}^{\infty}u^{-p}\,du\) converges exactly when \(p>1\). Thus the series converges for \(p>1\) and diverges for \(0<p\le1\). 3. If \(p=0\), every term equals \(1\). If \(p<0\), the terms grow as a positive power of \(n+c\). In both cases the terms do not approach \(0\), so the series diverges.

The series converges for \(p>1\) and diverges for \(p\le1\).

- Choose a continuous version of the summand and check the required tail behavior. - Which repeated inner expression could simplify the accumulated area? - Does the transformed tail approach a finite endpoint value?

1. Let \(f(x)=\frac{\sqrt x}{(x^{3/2}+1)^2}\). It is positive and continuous for \(x\ge1\). 2. Its derivative has the sign of \(1-5x^{3/2}\), so \(f\) is decreasing on \([1,\infty)\). 3. With \(u=x^{3/2}+1\) and \(du=\frac32\sqrt x\,dx\), \(\int_1^{\infty}f(x)\,dx=\frac23\int_2^{\infty}u^{-2}\,du=\frac13\). 4. The improper integral is finite, so the series converges by the integral test.

The series converges. The associated improper integral equals \(\frac13\).

- Study the long-term change of the continuous extension rather than only its first few values. - Which repeated logarithmic quantity makes the accumulated area simpler? - Does the final endpoint expression remain finite?

1. Let \(f(x)=\frac{\ln x}{x[1+(\ln x)^2]}\). It is positive and continuous for \(x\ge3\). 2. With \(t=\ln x\ge1\), the corresponding expression is \(\frac{t}{e^t(1+t^2)}\), whose derivative has numerator \(1-t-t^2-t^3<0\). Thus \(f\) is decreasing on the tail. 3. Substitute \(u=\ln x\), so \(\int_3^{\infty}f(x)\,dx=\int_{\ln3}^{\infty}\frac{u}{1+u^2}\,du\). 4. The antiderivative \(\frac12\ln(1+u^2)\) is unbounded, so the integral and series diverge.

The series diverges.

- Look for one denominator factor whose rate of change is represented by another factor. - What simpler variable is naturally suggested by that relationship? - What limiting value does the transformed endpoint approach?

1. Let \(f(x)=\frac{1}{(x^2+1)\arctan x}\). For \(x\ge1\), it is positive and continuous, and its denominator is increasing, so \(f\) is decreasing. 2. Use \(u=\arctan x\), so \(du=\frac{dx}{x^2+1}\). 3. Then \(\int_1^{\infty}f(x)\,dx=\int_{\pi/4}^{\pi/2}\frac{1}{u}\,du=\ln2\). 4. The integral is finite, so the series converges by the integral test.

The series converges. The associated improper integral equals \(\ln2\).

- A finite initial portion does not determine the long-term classification. - Find a convenient point after which the continuous terms move only downward. - Can the accumulated tail be reduced step by step to simpler endpoint expressions?

1. Let \(f(x)=x^3e^{-x}\). It is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=x^2e^{-x}(3-x)\), the function is decreasing for \(x\ge3\). Thus the tail beginning at \(n=4\) satisfies the integral-test hypotheses. 3. Repeated integration by parts gives \(\int_4^{\infty}x^3e^{-x}\,dx=e^{-4}(4^3+3\cdot4^2+6\cdot4+6)=\frac{142}{e^4}\). 4. This tail integral is finite, so the tail series converges. Adding the first three terms does not change convergence, so the original series converges.

The series converges. On the decreasing tail \([4,\infty)\), the associated improper integral equals \(\frac{142}{e^4}\).

- Check every hypothesis of the test, not only agreement at integer inputs. - Examine what the added oscillatory term does between consecutive integers. - Compare the invalid extension with a natural extension that preserves the required shape.

1. Although \(g(n)=1/n^2\), the function \(g\) is not decreasing on any tail because the term \(\sin^2(\pi x)\) continues to oscillate between \(0\) and \(1\). 2. Therefore \(g\) does not satisfy the hypotheses of the integral test. 3. A valid extension is \(f(x)=1/x^2\), which is positive, continuous, and decreasing on \([1,\infty)\). 4. Its improper integral is \(\int_1^{\infty}x^{-2}\,\mathrm{d}x=1\), so the integral test correctly confirms convergence. 5. Matching the sequence at integer inputs is not enough; the chosen extension must also satisfy the test's conditions.

Eero's extension is not decreasing, so the integral test cannot be applied to it. Using \(f(x)=1/x^2\) gives a convergent improper integral and confirms that the series converges.

- Use the same finite upper endpoint for both logarithmic terms. - What happens to their difference even though each term grows without bound? - Decide convergence only after taking the combined endpoint limit.

1. The function \(f(x)=1/[x(1+\sqrt x)]\) is positive, continuous, and decreasing for \(x\ge1\). 2. With \(u=\sqrt x\), the integral becomes \(2\int_1^{\infty}[1/u-1/(u+1)]\,\mathrm{d}u\). 3. The two divergent pieces cannot be evaluated separately; their difference must be integrated with a common finite upper endpoint. 4. For \(R>1\), the integral equals \(2[\ln u-\ln(u+1)]_1^R\). Taking \(R\to\infty\) gives \(2\ln2\). 5. The associated integral is finite, so the series converges.

Separating the two divergent improper integrals is invalid because their cancellation is lost. The combined integral equals \(2\ln2\), so the series converges.

- Factor the cubic denominator over the real numbers. - Separate the rational expression into terms with simpler denominators. - Check that the logarithmic growth cancels when the upper endpoint tends to infinity.

1. Let \(f(x)=\frac{1}{x^3+1}\). It is positive and continuous for \(x\ge1\), and \(f'(x)=-\frac{3x^2}{(x^3+1)^2}<0\). 2. Factor \(x^3+1=(x+1)(x^2-x+1)\) and write \(\frac{1}{x^3+1}=\frac{1}{3(x+1)}+\frac{-x+2}{3(x^2-x+1)}\). 3. An antiderivative is \(\frac13\ln(x+1)-\frac16\ln(x^2-x+1)+\frac{1}{\sqrt3}\arctan\!\left(\frac{2x-1}{\sqrt3}\right)\). 4. Evaluating from \(1\) to infinity gives \(\int_1^{\infty}f(x)\,dx=\frac{\pi}{3\sqrt3}-\frac13\ln2\), which is finite. Therefore the series converges.

The series converges. The associated improper integral equals \(\frac{\pi}{3\sqrt3}-\frac13\ln2\).

- The accumulated area need not be found exactly to determine its behavior. - Can the curved function be bounded below by a simpler line on the relevant input range? - What happens when that lower bound is accumulated over an infinite interval?

1. Let \(f(x)=\sin\!\left(\frac1x\right)\). It is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{\cos(1/x)}{x^2}<0\) on \([1,\infty)\), the function is decreasing. 3. The sine function is concave on \([0,1]\), so \(\sin y\ge y\sin1\) there. With \(y=1/x\), \(f(x)\ge\frac{\sin1}{x}\) for \(x\ge1\). 4. Thus \(\int_1^{\infty}f(x)\,dx\ge\sin1\int_1^{\infty}\frac{dx}{x}=\infty\). The series diverges by the integral test.

The series diverges.

- Replace the repeated logarithmic expression with one variable. - Factor the transformed denominator into its linear and quadratic parts. - Keep combined logarithms together when taking the infinite-endpoint limit.

1. Let \(f(x)=\frac{1}{x(1+\ln x)[1+(\ln x)^2]}\). It is positive and continuous for \(x\ge1\), and every positive denominator factor is increasing, so \(f\) is decreasing. 2. Set \(u=\ln x\). Then \(\int_1^{\infty}f(x)\,dx=\int_0^{\infty}\frac{du}{(u+1)(u^2+1)}\). 3. Decompose \(\frac{1}{(u+1)(u^2+1)}=\frac{1}{2(u+1)}+\frac{1-u}{2(u^2+1)}\). 4. The logarithmic terms cancel at infinity, and the inverse-tangent term gives \(\int_1^{\infty}f(x)\,dx=\frac{\pi}{4}\). Therefore the series converges.

The series converges. The associated improper integral equals \(\frac{\pi}{4}\).

- How can the unseen tail be bounded by the area under a related decreasing curve? - Translate the required error tolerance into a condition on the starting index. - Check the neighboring integers to ensure the chosen index is the least possible one.

1. Let \(f(x)=1/[x(\ln x)^2]\). It is positive, continuous, and decreasing for \(x\ge2\). 2. The improper integral is \(\int_2^{\infty}\frac{1}{x(\ln x)^2}\,dx=1/\ln2\), so the series converges by the integral test. 3. For the remainder \(R_N\), the integral-test bound gives \(R_N\le\int_N^{\infty}\frac{1}{x(\ln x)^2}\,dx=1/\ln N\). 4. To guarantee \(R_N\le0.2\), require \(1/\ln N\le0.2\), or \(\ln N\ge5\). 5. Thus \(N\ge e^5\), and the least integer satisfying the bound is \(N=149\).

The series converges. The integral-test remainder bound guarantees an error of at most \(0.2\) after summing through index \(N=149\).

- Use the first extension to obtain information about the common integer values. - Compare the area over each unit interval with the value at one endpoint. - Turn the interval-by-interval bounds into a bound for a finite accumulated area.

1. Because \(f\) is positive, continuous, and decreasing and its improper integral converges, the integral test implies that \(\sum_{n=1}^{\infty}a_n\) converges. 2. Since \(g\) is decreasing, \(g(x)\le g(n)=a_n\) for \(x\in[n,n+1]\). 3. Therefore \(\int_1^{N+1}g(x)\,dx\le\sum_{n=1}^{N}a_n\). 4. The partial sums on the right are bounded, so the increasing integrals on the left are bounded. Hence \(\int_1^{\infty}g(x)\,dx\) converges.

The second improper integral must converge because its finite upper-limit integrals are bounded above by partial sums of the convergent series \(\sum a_n\).

- Check eventual decrease as well as positivity and continuity; the sign of \(p\) should not be assumed. - A logarithmic derivative can make the monotonicity check shorter than differentiating the full quotient directly. - In the improper integral, look for a substitution whose differential matches the factor \(\frac{dx}{x\ln x}\).

1. Let \(f_p(x)=\frac{1}{x\ln x\,[\ln(\ln x)]^p}\). For \(x\ge3\), it is positive and continuous. 2. Its logarithmic derivative is \(\frac{f_p'(x)}{f_p(x)}=-\frac{1}{x}\left(1+\frac{1}{\ln x}+\frac{p}{\ln x\,\ln(\ln x)}\right)\). The expression in parentheses tends to \(1\), so it is positive for all sufficiently large \(x\). Thus \(f_p\) is eventually decreasing for every real \(p\). 3. With \(u=\ln(\ln x)\), so \(du=\frac{dx}{x\ln x}\), the associated improper integral becomes \(\int_{\ln(\ln3)}^{\infty}u^{-p}\,du\). 4. This power integral converges exactly when \(p>1\). Therefore the series converges for \(p>1\) and diverges for \(p\le1\).

The series converges for \(p>1\) and diverges for \(p\le1\).

- Differentiate the continuous extension to determine where it becomes decreasing. - Rewrite \(x^pe^{-x}\) as a single exponential involving \(p\ln x-x\). - Compare the tail with an exponential function whose improper integral is finite.

1. Let \(f_p(x)=x^pe^{-x}\). It is positive and continuous for \(x\ge1\). 2. Since \(f_p'(x)=x^{p-1}e^{-x}(p-x)\), the function is decreasing whenever \(x>p\); thus it is eventually decreasing for every real \(p\). 3. Because \(\frac{\ln x}{x}\to0\), there is an \(M\) such that \(p\ln x\le\frac{x}{2}\) for \(x\ge M\). Hence \(x^pe^{-x}=e^{p\ln x-x}\le e^{-x/2}\) on that tail. 4. The integral of \(e^{-x/2}\) over \([M,\infty)\) is finite, so \(\int_1^{\infty}x^pe^{-x}\,dx\) converges. Therefore the series converges for every real \(p\).

The series converges for every real \(p\).

- Find an endpoint expression whose change matches the continuous summand. - Use the areas beginning at two consecutive integer inputs to trap the total. - Include the opening series term separately in both bounds.

1. Let \(f(x)=(2x+1)/(x^2+x+1)^2\). It is positive and continuous for \(x\ge1\), and \(f'(x)=-6x(x+1)/(x^2+x+1)^3<0\). 2. Since \(\int_c^{\infty}f(x)\,dx=1/(c^2+c+1)\), the improper integral is finite, so the series converges. 3. For a positive decreasing function, \(a_1+\int_2^{\infty}f(x)\,dx\le S\le a_1+\int_1^{\infty}f(x)\,dx\). 4. Here \(a_1=1/3\), \(\int_2^{\infty}f(x)\,dx=1/7\), and \(\int_1^{\infty}f(x)\,dx=1/3\). 5. Therefore \(10/21\le S\le2/3\).

The series converges, and its sum satisfies \(\frac{10}{21}\le S\le\frac23\).

- Check the term behavior separately when the power in the denominator is nonpositive. - An exponential substitution converts powers of the index into exponential decay or growth. - Compare the transformed exponential rate with zero.

1. Let \(f_p(x)=(\ln x)^2/x^p\). It is positive and continuous for \(x\ge2\). 2. Its derivative is \(x^{-p-1}\ln x\,(2-p\ln x)\), so it is eventually decreasing when \(p>0\). If \(p\le0\), the terms do not approach \(0\), so the series diverges. 3. For \(p>0\), set \(x=e^u\). The integral becomes \(\int_{\ln2}^{\infty}u^2e^{-(p-1)u}\,du\). 4. This integral converges when \(p>1\) and diverges when \(0<p\le1\). Hence the series converges exactly when \(p>1\).

The series converges for \(p>1\) and diverges for \(p\le1\).

- Rewrite the quadratic so its center and positive remainder are visible. - Look for an endpoint expression whose rate of change matches the integrand. - Evaluate that expression at the finite and infinite endpoints.

1. Let \(f(x)=\frac{1}{(x^2+x+1)^{3/2}}\). It is positive and continuous for \(x\ge1\). 2. Since \(f'(x)=-\frac{3(2x+1)}{2(x^2+x+1)^{5/2}}<0\), the function is decreasing. 3. Completing the square gives an antiderivative \(\frac{2(2x+1)}{3\sqrt{x^2+x+1}}\). 4. Therefore \(\int_1^{\infty}f(x)\,dx=\frac43-\frac{2}{\sqrt3}\), which is finite. The series converges by the integral test.

The series converges. The associated improper integral equals \(\frac43-\frac{2}{\sqrt3}\).

- Compare a decreasing function with its endpoint values on one unit interval. - Integrate the pointwise inequalities before summing them. - Align the indices carefully with the first omitted term.

1. Since \(f\) is decreasing, on \([n,n+1]\) one has \(f(n+1)\le f(x)\le f(n)\). 2. Integrating gives \(f(n+1)\le\int_n^{n+1}f(x)\,dx\le f(n)\). 3. Sum the left inequalities for \(n=N,N+1,\ldots\) to obtain \(R_N\le\int_N^{\infty}f(x)\,dx\). 4. Sum the right inequalities beginning at \(n=N+1\) to obtain \(\int_{N+1}^{\infty}f(x)\,dx\le R_N\).

\(\int_{N+1}^{\infty}f(x)\,dx\le R_N\le\int_N^{\infty}f(x)\,dx\).