Free math worksheets

- Separate the alternating sign from the positive magnitude. - Does the magnitude approach zero? - Is the magnitude eventually moving in one direction?

1. The positive magnitude is \(b_n=\frac1n\). 2. \(b_n\to0\). 3. \(b_{n+1}<b_n\) for every \(n\ge1\). 4. Therefore the series converges by the alternating series test.

The magnitudes are \(b_n=1/n\). They decrease for \(n\ge1\) and satisfy \(b_n\to0\), so the series converges by the alternating series test.

- Remove the factor \((-1)^n\) to identify the positive magnitude \(1/(n+2)\). - Compare consecutive denominators to verify that the magnitudes decrease. - Use the denominator’s growth to check that the magnitudes approach \(0\).

1. The positive magnitude is \(b_n=\frac1{n+2}\). 2. \(b_n\to0\). 3. \(b_{n+1}<b_n\) for every \(n\ge1\). 4. Therefore the series converges by the alternating series test.

The magnitudes are \(b_n=1/(n+2)\). They decrease for \(n\ge1\) and satisfy \(b_n\to0\), so the series converges by the alternating series test.

- One condition concerns the long-term order of consecutive magnitudes. - The other condition concerns the magnitude limit. - Both conditions need only hold on a sufficiently late tail.

1. The positive magnitudes \(b_n\) must be nonincreasing for all sufficiently large \(n\). 2. They must satisfy \(\lim_{n\to\infty}b_n=0\). 3. Under these conditions, the alternating series converges.

The positive magnitudes \(b_n\) must be eventually nonincreasing and must approach \(0\).

- Check the necessary term limit before using monotonicity. - Account for the alternating sign when describing the term behavior. - Identify which required condition fails.

1. The terms of the series are \((-1)^n b_n\). 2. Since \(b_n\to L>0\), the series terms alternate near \(L\) and \(-L\) rather than approaching \(0\). 3. Therefore the series diverges by the \(n\)th-term test. 4. The decreasing condition does not compensate for failure of the zero-limit condition.

The series diverges because its terms do not approach \(0\).

- Use \(b_n=1/\sqrt n\) as the positive magnitude sequence. - The square-root function increases, so determine what that implies for its reciprocals. - Check the magnitude limit separately before invoking the alternating series test.

1. The positive magnitude is \(b_n=\frac1{\sqrt n}\). 2. \(b_n\to0\). 3. \(b_{n+1}<b_n\) for every \(n\ge1\). 4. Therefore the series converges by the alternating series test.

Yes. The magnitudes are \(b_n=1/\sqrt n\); they decrease for \(n\ge1\) and satisfy \(b_n\to0\). Therefore the series converges by the alternating series test.

- Remove the factor \((-1)^n\) to identify the positive magnitude sequence. - Determine what happens to \((2n+1)^2\) as \(n\) increases. - Compare consecutive denominators to verify monotonic decrease.

1. The positive magnitude is \(b_n=\frac1{(2n+1)^2}\). 2. \(b_n\to0\). 3. Because \((2n+3)^2>(2n+1)^2\), \(b_{n+1}<b_n\) for every \(n\ge1\). 4. Therefore the series converges by the alternating series test.

Here \(b_n=\frac1{(2n+1)^2}\), with \(b_n\to0\) and \(b_{n+1}<b_n\); therefore the series converges by the alternating series test.

- Treat \(\frac1n+\frac1{n^2}\) as the positive magnitude sequence. - Analyze the limit of each reciprocal-power term separately. - Use the fact that a sum of positive decreasing sequences is decreasing.

1. The positive magnitude is \(b_n=\frac1n+\frac1{n^2}\). 2. Both \(\frac1n\) and \(\frac1{n^2}\) approach \(0\), so \(b_n\to0\). 3. Both positive components decrease as \(n\) increases, so their sum \(b_n\) decreases. 4. Therefore the series converges by the alternating series test.

The magnitude \(b_n=\frac1n+\frac1{n^2}\) decreases to \(0\), so the series converges by the alternating series test.

- Rewrite \(\sqrt[3]{n}\) using a rational exponent. - Decide what happens to the reciprocal of \(n^{1/3}\) as \(n\to\infty\). - Use the increase of \(n^{1/3}\) to justify decrease of the magnitude sequence.

1. Since \(\sqrt[3]{n}=n^{1/3}\), the positive magnitude is \(b_n=\frac1{n^{1/3}}\). 2. Because the exponent is positive, \(b_n\to0\). 3. Since \(n^{1/3}\) increases, its reciprocal \(b_n\) decreases for every \(n\ge1\). 4. Therefore the series converges by the alternating series test.

The magnitude is \(b_n=n^{-1/3}\); it decreases to \(0\), so the series converges by the alternating series test.

- Remove the alternating sign to identify the magnitude sequence. - Track how \(\sqrt n+1\) changes as \(n\) increases. - Use the behavior of a positive reciprocal when its denominator grows without bound.

1. The positive magnitude is \(b_n=\frac1{\sqrt n+1}>0\). 2. Since \(\sqrt n+1\to\infty\), \(b_n\to0\). 3. The denominator increases with \(n\), so \(b_n\) decreases. 4. Therefore the series converges by the alternating series test.

The magnitudes are positive, decrease, and approach \(0\); therefore the series converges by the alternating series test.

- Alternating signs alone are not enough; identify the positive magnitude sequence. - Examine the denominator to determine the limit of \(\frac2{n+1}\). - Compare \(b_{n+1}\) with \(b_n\) to verify decrease.

1. The positive magnitude is \(b_n=\frac2{n+1}\). 2. Since \(n+1\to\infty\), \(b_n\to0\). 3. Because \(n+2>n+1\), \(b_{n+1}=\frac2{n+2}<\frac2{n+1}=b_n\) for every \(n\ge1\). 4. Therefore the series converges by the alternating series test.

The signs alternate, the magnitudes \(b_n=\frac2{n+1}\) decrease, and \(b_n\to0\); therefore the series converges by the alternating series test.

- Identify the positive magnitude \(b_n=\frac{n+1}{n+2}\). - Compute its limit using the leading coefficients. - If the magnitude does not approach \(0\), use the necessary condition for series convergence.

1. The positive magnitude is \(b_n=\frac{n+1}{n+2}\). 2. Dividing the numerator and denominator by \(n\) gives \(b_n\to1\), not \(0\). 3. Thus the alternating series test cannot establish convergence. In fact, the terms do not approach \(0\), so the series diverges by the \(n\)th-term test for divergence.

The magnitudes approach \(1\), not \(0\), so the series diverges by the \(n\)th-term test for divergence.

- Identify the magnitude sequence \(1+\frac1n\). - Compute its limit rather than relying only on its decrease. - A convergent series must have terms that approach \(0\).

1. The positive magnitude is \(b_n=1+\frac1n\). 2. Although \(b_n\) decreases, \(b_n\to1\), not \(0\). 3. The required zero-limit condition fails, so the terms of the series do not approach \(0\). 4. Therefore the series diverges by the \(n\)th-term test for divergence.

The magnitudes approach \(1\), not \(0\), so the series diverges.

- Remove the factor \((-1)^n\) and examine the rational magnitude. - Use the leading coefficients to find the magnitude limit. - Apply the necessary condition for convergence if that limit is nonzero.

1. The positive magnitude is \(b_n=\frac{n^2}{n^2+1}\). 2. Dividing the numerator and denominator by \(n^2\) gives \(b_n\to1\). 3. Thus the terms do not approach \(0\), so the alternating series test cannot establish convergence. 4. The series diverges by the \(n\)th-term test for divergence.

The magnitudes approach \(1\), so the terms do not approach \(0\) and the series diverges.

- Distinguish evidence about six terms from a statement about all sufficiently large indices. - Identify both magnitude conditions required by the alternating series test. - Ask whether a finite list determines the limit of an infinite sequence.

1. The list shows decrease for only six terms. 2. The alternating series test requires eventual decrease throughout the infinite tail and a limiting magnitude of \(0\). 3. A finite list alone proves neither condition for the full sequence.

No. The finite list does not establish eventual decrease or a zero limit for the infinite sequence.

- Use \(r>-1\) to determine the sign of \(n+r\) for all included indices. - Track how \((n+r)^2\) changes as \(n\) increases. - Find the limit of its reciprocal.

1. Because \(r>-1\), \(n+r>0\) for every \(n\ge1\). 2. The positive magnitudes \(b_n=\frac1{(n+r)^2}\) decrease as \(n\) increases. 3. Also, \(b_n\to0\). 4. Thus the series converges by the alternating series test for every \(r>-1\).

The series converges by the alternating series test for every \(r>-1\).

- Apply the alternating series test only to the tail where the conditions are known. - Separate the terms before \(n=50\) from that infinite tail. - Recall how finitely many initial terms affect convergence.

1. The tail beginning at \(n=50\) satisfies the alternating series test. 2. The first \(49\) terms form a finite sum. 3. Adding or removing finitely many terms does not change convergence, so the full series converges.

Yes. The tail converges by the alternating series test, and the first \(49\) terms do not affect convergence.

- Check the zero-limit condition for each magnitude sequence. - Check whether each sequence is eventually nonincreasing. - A failed test hypothesis means the test is inconclusive, not necessarily that the series diverges.

1. The series with magnitudes \(b_n\) satisfies both alternating-series conditions, so it converges. 2. The series with magnitudes \(c_n\) fails the eventual-monotonicity condition. 3. Therefore the alternating series test is inconclusive for the second series.

The \(b_n\) series converges by the test; the test is inconclusive for the \(c_n\) series.

- Identify how the input to the inverse-trigonometric function changes with the index. - Use monotonicity of the outer function to transfer that change to the magnitudes. - Evaluate the magnitude limit by continuity.

1. Let \(b_n=\arctan(1/n)>0\). 2. Since \(1/(n+1)<1/n\) and \(\arctan x\) is increasing, \(b_{n+1}<b_n\). 3. Because \(1/n\to0\) and \(\arctan x\) is continuous at \(0\), \(b_n\to0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Check the nested logarithms from the inside outward. - Use monotonicity of each nested function to determine the direction of the reciprocal. - Evaluate whether the full denominator grows without bound.

1. Let \(b_n=1/\ln(\ln(n+16))\). For \(n\ge1\), \(n+16\ge17\), so the denominator is positive and the magnitudes are defined. 2. The functions \(n+16\), \(\ln(n+16)\), and \(\ln(\ln(n+16))\) all increase, so \((b_n)\) decreases. 3. Since \(\ln(\ln(n+16))\to\infty\), \(b_n\to0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Separate the finite irregular beginning from the infinite tail. - Check the required magnitude conditions only where the regular pattern begins. - Remember that finitely many initial terms cannot change convergence.

1. The first two terms form a finite initial sum and do not affect convergence. 2. For \(n\ge3\), the magnitudes are \(b_n=1/n\), which decrease to \(0\). 3. Therefore the tail beginning at \(n=3\) converges by the alternating series test. 4. Adding the two initial terms preserves convergence of the full series.

The series converges; the finite initial increase does not affect the alternating tail.

- Track how the inner expression changes with the index. - Use the monotonicity of the outer exponential function. - Evaluate the exponent's behavior to obtain the magnitude limit.

1. Let \(b_n=e^{-\sqrt n}>0\). 2. Since \(\sqrt n\) increases and \(e^{-x}\) decreases, the sequence \((b_n)\) decreases. 3. Because \(\sqrt n\to\infty\), \(e^{-\sqrt n}\to0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Use the given product limit to recover the limit of the magnitude itself. - Separate the bounded numerator from the growing index. - Combine the resulting zero limit with the stated monotonicity.

1. Since \(n b_n\to5\), write \(b_n=(n b_n)/n\). 2. The numerator approaches \(5\) while the denominator grows without bound, so \(b_n\to0\). 3. The magnitudes are given to be positive and decreasing. 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Track the input of the logarithm as the index increases. - Use monotonicity and continuity of the outer function. - Verify both magnitude conditions separately.

1. Let \(b_n=\ln(1+1/n)>0\). 2. Since \(1+1/n\) decreases and logarithm is increasing, \((b_n)\) decreases. 3. Because \(1+1/n\to1\), \(b_n\to\ln1=0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- One required condition is already stated directly. - Use the upper bound to determine the magnitude limit. - The lower bound is not needed for convergence but is consistent with positivity.

1. The sequence is given to be positive and nonincreasing. 2. The upper bound \(1/\sqrt n\to0\). 3. Since \(0<b_n\le1/\sqrt n\), the squeeze theorem gives \(b_n\to0\). 4. Therefore the alternating series converges by the alternating series test.

The series converges by the alternating series test.

- Compare the parity of an integer with the parity of its square. - Rewrite the sign pattern in the standard alternating form. - Check the reciprocal-power magnitudes against the two required conditions.

1. The integers \(n^2\) and \(n\) have the same parity, so \((-1)^{n^2}=(-1)^n\). 2. The positive magnitudes are \(b_n=1/n^{2/3}\). 3. The sequence \((b_n)\) decreases and approaches \(0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Treat the magnitude as the real-variable function \(f(x)=\frac{\ln x}{x}\). - Differentiate and translate the condition \(f'(x)<0\) into the first valid integer index. - Verify that the magnitudes approach \(0\), then apply the alternating series test.

1. The positive magnitude is \(b_n=\frac{\ln n}{n}\). 2. \(b_n\to0\). 3. For \(f(x)=\frac{\ln x}{x}\), \(f'(x)=\frac{1-\ln x}{x^2}<0\) when \(x>e\), so \(b_n\) decreases for every integer \(n\ge3\). 4. Therefore the series converges by the alternating series test.

The magnitudes decrease for \(n\ge3\) and approach \(0\), so the series converges by the alternating series test.

- Use \(b_n=\frac{n}{n^2+1}\) as the positive magnitude sequence. - Check the limit by comparing the powers of \(n\) in the numerator and denominator. - Differentiate \(f(x)=\frac{x}{x^2+1}\) to justify decrease before applying the alternating series test.

1. The positive magnitude is \(b_n=\frac{n}{n^2+1}\). 2. Dividing the numerator and denominator by \(n^2\) shows that \(b_n\to0\). 3. For \(f(x)=\frac{x}{x^2+1}\), \(f'(x)=\frac{1-x^2}{(x^2+1)^2}\le0\) for \(x\ge1\), so \(b_n\) is decreasing. 4. Therefore the series converges by the alternating series test.

The magnitudes decrease for \(n\ge1\) and approach \(0\), so the series converges by the alternating series test.

- Use \(b_n=\frac1{\ln(n+1)}\) for the positive magnitudes. - Recall the long-term behavior of \(\ln(n+1)\). - Decide what taking the reciprocal of a positive increasing function does to monotonicity.

1. The positive magnitude is \(b_n=\frac1{\ln(n+1)}\). 2. Since \(\ln(n+1)\to\infty\), \(b_n\to0\). 3. Since \(\ln(n+1)\) is positive and increasing for \(n\ge1\), its reciprocal \(b_n\) decreases. 4. Therefore the series converges by the alternating series test.

The magnitudes are positive, decrease, and approach \(0\), so the series converges by the alternating series test.

- Set \(b_n=\frac{n}{n^2+4}\) and first determine its limit. - Differentiate \(f(x)=\frac{x}{x^2+4}\) and solve the inequality \(f'(x)\le0\). - The alternating series test requires decrease only eventually, not necessarily from the first term.

1. The positive magnitude is \(b_n=\frac{n}{n^2+4}\). 2. Dividing the numerator and denominator by \(n^2\) shows that \(b_n\to0\). 3. For \(f(x)=\frac{x}{x^2+4}\), \(f'(x)=\frac{4-x^2}{(x^2+4)^2}\le0\) for \(x\ge2\), so \(b_n\) decreases for \(n\ge2\). 4. Therefore the series converges by the alternating series test.

The magnitudes decrease for \(n\ge2\) and approach \(0\), so the series converges by the alternating series test.

- Use \(b_n=\frac{n+5}{(n+1)^2}\) and compare the degrees to find its limit. - Differentiate the continuous extension \(f(x)=\frac{x+5}{(x+1)^2}\). - Check the derivative sign starting at \(x=1\) to decide whether any initial terms are exceptions.

1. The positive magnitude is \(b_n=\frac{n+5}{(n+1)^2}\). 2. Comparing the leading powers shows that \(b_n\to0\). 3. For \(f(x)=\frac{x+5}{(x+1)^2}\), \(f'(x)=\frac{-x-9}{(x+1)^3}<0\) for \(x\ge1\), so there are no initial exceptions to decrease. 4. Therefore the series converges by the alternating series test.

The magnitudes decrease from \(n=1\) onward and approach \(0\), so the series converges by the alternating series test; there are no finite initial exceptions.

- Compare the highest powers in \(b_n=\frac{n+1}{n^2+1}\) to determine its limit. - Differentiate the corresponding function and simplify the numerator of the derivative. - Verify where that numerator is negative before applying the alternating series test.

1. The positive magnitude is \(b_n=\frac{n+1}{n^2+1}\). 2. Comparing the leading powers shows that \(b_n\to0\). 3. For \(f(x)=\frac{x+1}{x^2+1}\), \(f'(x)=\frac{-x^2-2x+1}{(x^2+1)^2}<0\) for \(x\ge1\), so \(b_n\) decreases from the first term onward. 4. Therefore the series converges by the alternating series test.

The magnitudes approach \(0\) and decrease for \(n\ge1\), so the series converges by the alternating series test.

- Extend the magnitudes to \(f(x)=\frac{\ln(x+1)}{x+1}\). - Differentiate and determine when \(1-\ln(x+1)\) is negative. - Compare that threshold with the series starting index \(n=2\).

1. The positive magnitude is \(b_n=\frac{\ln(n+1)}{n+1}\), and \(b_n\to0\). 2. Let \(f(x)=\frac{\ln(x+1)}{x+1}\). 3. Then \(f'(x)=\frac{1-\ln(x+1)}{(x+1)^2}<0\) when \(x+1>e\). Since every included integer satisfies \(n\ge2\), the magnitudes decrease throughout the series. 4. Therefore the series converges by the alternating series test.

The magnitudes decrease for all included indices \(n\ge2\) and approach \(0\), so the series converges by the alternating series test.

- Use \(b_n=\frac{n}{n^2+9}\) and first determine its limit. - Differentiate \(f(x)=\frac{x}{x^2+9}\). - Solve \(f'(x)\le0\) and translate the result into an integer threshold.

1. The positive magnitude is \(b_n=\frac{n}{n^2+9}\). 2. Comparing the leading powers shows that \(b_n\to0\). 3. For \(f(x)=\frac{x}{x^2+9}\), \(f'(x)=\frac{9-x^2}{(x^2+9)^2}\le0\) for \(x\ge3\), so \(b_n\) decreases for \(n\ge3\). 4. Therefore the series converges by the alternating series test.

The magnitudes decrease for \(n\ge3\) and approach \(0\), so the series converges by the alternating series test.

- For the alternating-series check, study \(b_n=\frac1{n^2+n}\). - Remember that the alternating series test establishes convergence, not automatically absolute convergence. - To test absolute convergence here, compare \(\frac1{n^2+n}\) with a reciprocal-square series.

1. The positive magnitude is \(b_n=\frac1{n^2+n}\). 2. Since the denominator grows without bound, \(b_n\to0\), and because the denominator increases, \(b_n\) decreases. 3. Therefore the series converges by the alternating series test. 4. This test alone does not establish absolute convergence. Here, however, \(0<\frac1{n^2+n}\le\frac1{n^2}\), so the series of absolute values converges by comparison with the convergent \(p\)-series \(\sum1/n^2\).

The alternating series test proves convergence. A separate comparison shows that the series is actually absolutely convergent.

- Compare degrees to determine the limit of the rational magnitude. - Differentiate the continuous extension and focus on the quadratic numerator of the derivative. - Find an integer threshold beyond its positive root.

1. The positive magnitude is \(b_n=\frac{3n+1}{n^2+4n+5}\). 2. Comparing the leading powers shows that \(b_n\to0\). 3. For \(f(x)=\frac{3x+1}{x^2+4x+5}\), \(f'(x)=\frac{-3x^2-2x+11}{(x^2+4x+5)^2}<0\) for \(x\ge2\), so \(b_n\) decreases for \(n\ge2\). 4. Therefore the series converges by the alternating series test.

The magnitudes decrease for \(n\ge2\) and approach \(0\), so the series converges by the alternating series test.

- Evaluate \(\lfloor n/2\rfloor\) for the first several positive integers. - Write the resulting signs in order and look for repeated signs. - Compare that pattern with the term-by-term sign change required by the alternating series test.

1. Listing the signs gives \(+,-,-,+,+,-,-,\ldots\). 2. After the first term, each sign occurs twice, so the sign does not change at every successive index. 3. Therefore the series is not in the standard alternating form required by the alternating series test, and the test does not apply directly.

The alternating series test does not apply directly because the signs occur in pairs rather than alternating term by term.

- Compute the magnitude separately for odd and even indices. - Compare \(b_{2k-1}\) with \(b_{2k}\). - A zero magnitude limit does not replace the test’s eventual-decrease condition.

1. The positive magnitudes are \(b_n=\frac{2+(-1)^n}{n}\), and \(b_n\to0\). 2. For every \(k\ge1\), \(b_{2k-1}=\frac1{2k-1}\) and \(b_{2k}=\frac3{2k}\), so \(b_{2k}>b_{2k-1}\). 3. Thus the magnitudes are not eventually decreasing. 4. The alternating series test is inconclusive.

The alternating series test is inconclusive because the magnitudes are not eventually decreasing.

- Separate the cases according to the sign of \(p\). - Determine when \((2n+1)^{-p}\) approaches \(0\). - In the zero-limit cases, check whether the magnitude decreases as \(n\) increases.

1. If \(p>0\), the positive magnitudes \(b_n=(2n+1)^{-p}\) decrease and approach \(0\). 2. Therefore the series converges by the alternating series test when \(p>0\). 3. If \(p=0\), then \(b_n=1\); if \(p<0\), the magnitudes grow without bound. 4. Thus the terms fail to approach \(0\) when \(p\le0\), so the series diverges in those cases.

The series converges by the alternating series test for \(p>0\) and diverges for \(p\le0\).

- Rewrite the magnitude as \(n^{-p}\) and consider the sign of \(p\). - Identify exactly when the magnitude approaches \(0\). - Use the alternating series test in the zero-limit range and the \(n\)th-term test for divergence otherwise.

1. If \(p>0\), the magnitudes \(1/n^p\) decrease to \(0\), so the series converges by the alternating series test. 2. If \(p=0\), the terms are \(\pm1\). 3. If \(p<0\), the magnitudes \(n^{-p}\) grow rather than approach \(0\). 4. Therefore the series diverges by the \(n\)th-term test for divergence when \(p\le0\).

The series converges for \(p>0\) and diverges for \(p\le0\).

- Find the limit of \(\frac{n}{n^2+a}\) for a fixed positive \(a\). - Differentiate the continuous extension \(\frac{x}{x^2+a}\). - Solve the derivative-sign inequality to show that decrease begins after a finite threshold.

1. The magnitude \(b_n=\frac{n}{n^2+a}\) approaches \(0\). 2. For \(f(x)=\frac{x}{x^2+a}\), \(f'(x)=\frac{a-x^2}{(x^2+a)^2}\). 3. Thus \(f'(x)<0\) when \(x>\sqrt a\), so the magnitudes are eventually decreasing for every \(a>0\). 4. Therefore the series converges by the alternating series test for every \(a>0\).

The series converges by the alternating series test for every \(a>0\).

- Isolate the positive magnitude sequence. - Use a continuous extension to check its direction of change. - Rewrite the magnitude so its limit is clear.

1. Let \(b_n=\frac{\sqrt n}{n+1}\). The magnitudes are positive. 2. For \(f(x)=\frac{\sqrt x}{x+1}\), \(f'(x)=\frac{1-x}{2\sqrt x(x+1)^2}\le0\) for \(x\ge1\). Thus \((b_n)\) is nonincreasing. 3. Dividing numerator and denominator by \(n\) gives \(b_n=\frac{1/\sqrt n}{1+1/n}\to0\). 4. Therefore the alternating series converges by the alternating series test.

The series converges by the alternating series test.

- Compare two consecutive tails rather than trying to evaluate either tail exactly. - Recall what happens to the remainders of a convergent positive series. - Check the two magnitude conditions separately.

1. Each \(b_n\) is positive because it is a tail of a positive series. 2. Since \(b_n-b_{n+1}=1/n^2>0\), the sequence \((b_n)\) is decreasing. 3. The series \(\sum 1/k^2\) converges, so its tails satisfy \(b_n\to0\). 4. Therefore \(\sum(-1)^n b_n\) converges by the alternating series test.

The series converges by the alternating series test.

- Rewrite the difference of radicals in a form with a denominator. - Study how that denominator changes with the index. - Use the rewritten form to evaluate the limit.

1. Let \(b_n=\sqrt{n+1}-\sqrt n>0\). 2. Rationalizing gives \(b_n=\frac{1}{\sqrt{n+1}+\sqrt n}\). 3. The denominator increases, so \((b_n)\) decreases, and the denominator tends to infinity, so \(b_n\to0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Strict decrease is not required; repeated magnitudes are allowed. - Examine the contribution of each consecutive pair. - Compare even and odd partial sums to identify the total.

1. The magnitude sequence is \(1,1,1/2,1/2,1/3,1/3,\ldots\), so it is nonincreasing and approaches \(0\). 2. The alternating series test therefore proves convergence. 3. Consecutive pairs cancel: \(b_{2k-1}-b_{2k}=0\). 4. Every even partial sum is \(0\), and odd partial sums equal \(1/k\to0\), so the sum is \(0\).

The series converges to \(0\).

- Track the distance between the recursive term and its fixed value. - Determine the limiting magnitude before considering the signs. - Check the necessary zero-term condition for any convergent series.

1. Rewrite the recurrence as \(1-b_{n+1}=\frac12(1-b_n)\). 2. Since \(1-b_1=1/2\), induction gives \(1-b_n=2^{-n}\), so \(b_n=1-2^{-n}\). 3. Therefore \(b_n\to1\), not \(0\). 4. The terms \((-1)^n b_n\) do not approach \(0\), so the series diverges by the \(n\)th-term test. 5. The alternating series test cannot prove convergence because its zero-magnitude condition fails.

The series diverges because \(b_n\to1\), so its terms do not approach \(0\).

- Compare the integrands for consecutive indices on the interval of integration. - Bound the denominator in a direction that produces a familiar integral. - Verify positivity, decrease, and the limiting value of the magnitudes.

1. The integrand is positive on \([0,1]\), so \(b_n>0\). 2. For \(0<x<1\), \(x^{n+1}<x^n\). Integrating after division by \(1+x>0\) gives \(b_{n+1}<b_n\). 3. Also \(0<b_n\le\int_0^1x^n\,dx=1/(n+1)\to0\). 4. The magnitudes are decreasing and approach \(0\), so the alternating series converges.

The series converges by the alternating series test.

- Use positivity of the integrand to establish positive magnitudes. - Compare integrals over adjacent intervals using how the integrand changes. - Bound each unit-interval area by a rectangle whose height tends to zero.

1. Each \(b_n\) is positive because the integrand is positive. 2. Since \(e^{-x^2}\) decreases for \(x>0\), shifting the unit interval to the right decreases its integral, so \(b_{n+1}<b_n\). 3. On \([n,n+1]\), \(e^{-x^2}\le e^{-n^2}\), so \(0<b_n\le e^{-n^2}\to0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Determine whether the rounded denominator can ever decrease. - Remember that nonincreasing magnitudes may contain plateaus. - Use unbounded denominator growth to evaluate the limit.

1. Let \(b_n=1/\lceil\sqrt n\rceil>0\). 2. The sequence \(\lceil\sqrt n\rceil\) is nondecreasing, so \((b_n)\) is nonincreasing, with repeated equal values allowed. 3. Since \(\lceil\sqrt n\rceil\to\infty\), \(b_n\to0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Confirm the difference is positive using monotonicity of the outer function. - Rewrite the difference as one inverse-trigonometric expression. - Track the new input as the index grows.

1. The magnitudes are positive because \(\arctan x\) is increasing. 2. The subtraction formula gives \(\arctan(n+1)-\arctan n=\arctan\!\left(\frac{1}{n^2+n+1}\right)\). 3. The inner fraction decreases to \(0\), so the magnitudes decrease and approach \(0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- Simplify the sign expression separately for even and odd indices. - Inspect the signs of the nonzero terms rather than the written appearance. - Reindex the surviving subseries.

1. For even \(n\), \((-1)^n-1=0\), so the term is \(0\). 2. For odd \(n\), \((-1)^n-1=-2\), so the term is \(-1/n\). 3. The nonzero terms are all negative; they do not alternate in sign. 4. Their magnitudes form the odd harmonic subseries, which diverges, so the given series diverges to \(-\infty\).

The series diverges to \(-\infty\); its nonzero terms do not alternate.

- Compare the intervals of integration for consecutive indices. - Bound the integrand by a simple constant on those intervals. - Use the interval length to determine the limiting magnitude.

1. Each \(b_n>0\) because the integrand is positive. 2. The intervals are nested: \([0,1/(n+1)]\subset[0,1/n]\). Therefore \(b_{n+1}<b_n\). 3. Since \(0<e^{-x^2}\le1\), \(0<b_n\le\int_0^{1/n}1\,dx=\frac1n\to0\). 4. The magnitudes are positive, decreasing, and approach \(0\), so the series converges by the alternating series test.

The series converges by the alternating series test.

- Apply the test to the portion where all its conditions hold. - Separate the finite initial segment from the infinite tail. - Finitely many exceptions cannot change convergence.

1. The tail \(\sum_{n=N}^{\infty}(-1)^n b_n\) has positive magnitudes that are nonincreasing and approach \(0\). 2. The alternating series test gives convergence of that tail. 3. The terms before index \(N\) form a finite sum. 4. Adding a finite initial sum to a convergent tail preserves convergence.

The series converges.

- Determine how the partial harmonic sums change with the index. - Establish that their growth is unbounded using grouped terms. - Transfer those properties through reciprocals.

1. The harmonic numbers are positive and strictly increasing, so \(b_n=1/H_n\) is positive and decreasing. 2. The harmonic numbers are unbounded; for example, \(H_{2^m}\ge1+m/2\). 3. Hence \(H_n\to\infty\) and \(b_n\to0\). 4. The series converges by the alternating series test.

The series converges by the alternating series test.

- First determine for which \(q\) the power \(n^q\) dominates \(\ln n\). - Differentiate \(f(x)=\frac{\ln x}{x^q}\) to locate an eventual-decrease threshold. - For the remaining parameter values, use the necessary zero-term condition.

1. If \(q>0\), then \(b_n=\frac{\ln n}{n^q}\to0\). 2. For \(f(x)=\frac{\ln x}{x^q}\), \(f'(x)=x^{-q-1}(1-q\ln x)\), which is negative when \(x>e^{1/q}\). Thus the magnitudes are eventually decreasing. 3. Therefore the series converges by the alternating series test for \(q>0\). 4. If \(q=0\), the magnitudes equal \(\ln n\); if \(q<0\), they grow even faster. In both cases the terms fail to approach \(0\), so the series diverges.

The series converges by the alternating series test for \(q>0\) and diverges for \(q\le0\).

- A fixed value of \(c\) affects only finitely many signs of \(n+c\). - Compare the degrees to determine the limit for every fixed \(c\). - In the derivative numerator, identify which term dominates as \(x\to\infty\).

1. For every fixed real \(c\), \(\frac{n+c}{n^2+1}\to0\). 2. The numerator \(n+c\) is positive for all sufficiently large \(n\), so the eventual positive magnitude is \(b_n=\frac{n+c}{n^2+1}\). 3. For \(f(x)=\frac{x+c}{x^2+1}\), \(f'(x)=\frac{-x^2-2cx+1}{(x^2+1)^2}\), which is negative for all sufficiently large \(x\). 4. Thus the magnitude is eventually positive, decreasing, and approaching \(0\), so the series converges by the alternating series test for every real \(c\).

The series satisfies the alternating series test eventually and converges for every real \(c\).

- Show that the recurrence preserves a useful positive interval. - Determine whether each new magnitude is smaller than the preceding one. - Use the recurrence again after the magnitude sequence is known to have a limit.

1. If \(0<b_n\le1\), then \(b_{n+1}=b_n(1-b_n/2)\) is positive and less than \(b_n\). By induction, \((b_n)\) is positive and decreasing. 2. Therefore \((b_n)\) has a limit \(L\ge0\). 3. Passing to the limit in the recurrence gives \(L=L-L^2/2\), so \(L=0\). 4. The magnitudes decrease to \(0\), so the alternating series converges.

The series converges by the alternating series test.

- Check the zero-limit condition using one bound that covers both cases. - Compare the magnitude immediately before a perfect square with the magnitude at the square. - Decide whether the required monotonic behavior eventually begins.

1. The magnitudes are positive, and \(0<b_n\le1/\sqrt n\), so \(b_n\to0\). 2. For every integer \(m\ge2\), \(b_{m^2-1}=1/(m^2-1)\) while \(b_{m^2}=1/m\). 3. Thus \(b_{m^2}>b_{m^2-1}\) for infinitely many indices, so the magnitudes are not eventually nonincreasing. 4. The alternating series test is therefore inconclusive for this series.

The alternating series test is inconclusive because the magnitudes approach \(0\) but increase at infinitely many perfect-square indices.

- Examine the net contribution of each consecutive pair. - Compare the positive part of a pair with a divergent benchmark. - Identify which condition besides a zero limit is needed for the standard alternating test.

1. Group consecutive terms: \((-1)^{2k-1}b_{2k-1}+(-1)^{2k}b_{2k} =-\frac1k+\frac{1}{\sqrt k}\). 2. For \(k\ge4\), \(1/k\le1/(2\sqrt k)\), so \(\frac{1}{\sqrt k}-\frac1k\ge\frac{1}{2\sqrt k}\). 3. The series of paired terms diverges to \(+\infty\) by comparison with \(\sum1/\sqrt k\). 4. Therefore the original series diverges. 5. Although \(b_n\to0\), the magnitudes are not eventually nonincreasing because \(b_{2k}=1/\sqrt k>b_{2k-1}=1/k\) for every \(k>1\); the required monotonicity condition is absent.

The series diverges. The magnitudes approach \(0\), but they are not eventually nonincreasing, so the alternating series test does not apply.

- Rewrite the magnitude to reveal its net power for the limit. - Use a continuous extension to determine eventual monotonicity. - Compare the parameter with the value that makes the net exponent zero.

1. Let \(b_n=n^p/(n+1)>0\). Its continuous extension satisfies \(f'(x)=\frac{x^{p-1}[(p-1)x+p]}{(x+1)^2}\). 2. If \(p<1\), the bracket is negative for all sufficiently large \(x\), so the magnitudes eventually decrease. 3. Also \(b_n=\frac{n^{p-1}}{1+1/n}\to0\) exactly when \(p<1\). 4. For \(p\ge1\), the magnitudes do not approach \(0\). Therefore the series converges by the alternating series test exactly when \(p<1\).

The series converges by the alternating series test exactly when \(p<1\), and diverges when \(p\ge1\).

- What does each recurrence multiplier imply about movement of the magnitudes? - Can the product of all multipliers be simplified through cancellation? - Does the resulting magnitude limit satisfy the necessary condition?

1. Each multiplier lies between \(0\) and \(1\), so \((b_n)\) is positive and decreasing. 2. Expanding the recurrence gives \(b_n=\prod_{j=3}^{n+1}\left(1-\frac{1}{j^2}\right)\). 3. The product telescopes: \(b_n=\frac{2}{n+1}\cdot\frac{n+2}{3}=\frac{2(n+2)}{3(n+1)}\to\frac23\). 4. The magnitudes do not approach \(0\), so the series diverges by the \(n\)th-term test.

The series diverges because \(b_n\to\frac23\), even though the magnitudes decrease.

- Compare the two magnitudes within each sign pair. - Compute the net contribution of a complete pair. - Control the difference between a paired partial sum and the next partial sum.

1. The magnitudes increase from \(b_{2k-1}\) to \(b_{2k}\) for every \(k\), so they are not eventually nonincreasing. 2. The \(k\)th signed pair is \(b_{2k-1}-b_{2k}=-1/k^2\). 3. The series of pair sums converges because \(\sum1/k^2\) converges. 4. The unpaired term in an odd partial sum is \(b_{2k-1}=1/k\to0\), so the full partial sums converge. Thus the alternating series test is sufficient but not necessary.

The series converges, but not by a direct application of the alternating series test because its magnitudes increase infinitely often.

- Use the ratio bound first to establish eventual decrease. - Apply the same bound repeatedly to control later magnitudes. - Compare the resulting bound with a decaying geometric sequence.

1. For \(n\ge N\), the ratio bound gives \(b_{n+1}\le rb_n<b_n\), so the magnitudes decrease on the tail. 2. Repeated use of the bound gives \(b_{N+k}\le b_Nr^k\). 3. Since \(0<r<1\), \(b_Nr^k\to0\), so \(b_n\to0\). 4. The alternating tail converges by the alternating series test, and finitely many initial terms do not affect convergence.

The series converges by the alternating series test.

- Place an arbitrary large index between consecutive members of the known subsequence. - Use monotonicity to trap the unknown magnitude between simpler quantities. - What does that bound imply for the full magnitude sequence?

1. For any sufficiently large \(n\), choose \(k\) so that \(2^k\le n<2^{k+1}\). 2. Since \((b_n)\) is nonincreasing, \(0<b_n\le b_{2^k}\). 3. As \(n\to\infty\), the corresponding \(k\to\infty\), and \(b_{2^k}\to0\). Thus \(b_n\to0\) by squeezing. 4. The magnitudes are positive, nonincreasing, and approach \(0\), so the alternating series converges.

The series converges by the alternating series test.

- Write the next harmonic number in terms of the current one. - Cross-multiply the comparison of consecutive magnitudes carefully. - Use a standard upper estimate for harmonic numbers to obtain the limit.

1. Let \(b_n=H_n/n>0\). 2. The inequality \(b_{n+1}\le b_n\) is equivalent to \(\frac{n}{n+1}\le H_n\), which holds because \(H_n\ge1\). 3. Since \(H_n\le1+\ln n\), \(0<b_n\le\frac{1+\ln n}{n}\to0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test.

- What happens when the recurrence is written in terms of reciprocals? - Can the resulting simpler recurrence be continued from the initial value? - Use the explicit magnitude sequence to check the required properties.

1. Taking reciprocals in the recurrence gives \(\frac{1}{b_{n+1}}=\frac{1}{b_n}+1\). 2. Since \(1/b_1=1\), induction gives \(1/b_n=n\), so \(b_n=1/n\). 3. The magnitudes are positive, decreasing, and approach \(0\). 4. Therefore the series converges by the alternating series test.

The series converges by the alternating series test; in fact, \(b_n=1/n\).

- Write out the first several signs before choosing a test. - Combine one complete sign cycle and simplify its net contribution. - Check the series of absolute values separately for the final classification.

1. The signs occur in pairs: \(+,+,-,-,+,+,-,-,\ldots\), so they do not alternate term by term. 2. Group four consecutive terms. For \(k\ge0\), the block sum is \(B_k=\frac1{4k+1}+\frac1{4k+2}-\frac1{4k+3}-\frac1{4k+4}\). 3. Rewrite \(B_k=\frac{2}{(4k+1)(4k+3)}+\frac{2}{(4k+2)(4k+4)}\), so \(0<B_k\le4/(4k+1)^2\). Hence the block-sum series converges. 4. Terms inside an incomplete final block approach \(0\), so the original series converges. Its series of absolute values is harmonic, so convergence is conditional.

The series converges conditionally. The ordinary alternating series test does not apply because the signs occur in pairs.

- Separate eventual positivity, the zero limit, and one-step decrease. - Compare the fixed-sign part of a consecutive difference with its parity-dependent part. - For the second part, rewrite the signed term as a sum of two familiar series.

1. For every \(c\), \(b_n\to0\), and \(b_n>0\) for all sufficiently large \(n\). 2. The consecutive difference is \(b_n-b_{n+1}=\frac{1}{n(n+1)}+c(-1)^n\left(\frac1{n^2}+\frac1{(n+1)^2}\right)\). 3. Let \(R_n=\frac{1/[n(n+1)]}{1/n^2+1/(n+1)^2} =\frac{n(n+1)}{2n^2+2n+1}\). Then \(R_n\to\frac12\) and \(R_n<\frac12\). 4. If \(|c|<\frac12\), then \(R_n>|c|\) for all sufficiently large \(n\), so \(b_n-b_{n+1}>0\) eventually. The alternating series test applies. 5. If \(|c|\ge\frac12\), choose the infinitely many indices for which \(c(-1)^n=-|c|\). At those indices, \(b_n-b_{n+1}\le\frac{1}{n(n+1)}-\frac12\left(\frac1{n^2}+\frac1{(n+1)^2}\right)<0\). Thus the magnitudes are not eventually nonincreasing, so the test cannot be applied directly. 6. Regardless of \(c\), \((-1)^n b_n=\frac{(-1)^n}{n}+\frac{c}{n^2}\). Both series on the right converge, so the given series converges for every real \(c\).

a) The alternating series test applies directly exactly when \(|c|<\frac12\). b) The series converges for every real \(c\), including the values for which that test is not directly applicable.