Free math worksheets

- Examine the term values for even and odd indices. - Decide whether the term sequence approaches \(0\) or continues switching between fixed values. - The \(n\)th-term test proves divergence whenever the term limit is nonzero or does not exist.

1. The term sequence \(a_n=(-1)^n\) has no limit. 2. In particular, \(a_n\) does not approach \(0\). 3. Therefore the series diverges.

The series diverges by the \(n\)th-term test.

- Determine what happens to \(\sqrt n\) and therefore to its reciprocal as \(n\to\infty\). - The \(n\)th-term test proves divergence only when the term limit is nonzero or nonexistent. - Do not treat the necessary condition \(a_n\to0\) as a sufficient condition for convergence.

1. \(\lim_{n\to\infty}a_n=0\). 2. A zero term limit is necessary for convergence but is not sufficient. 3. The \(n\)th-term test is inconclusive.

The general term approaches \(0\), so the \(n\)th-term test is inconclusive.

- Rewrite \(2^{-n}\) as \(\left(\frac12\right)^n\). - Determine the limit of this geometric term sequence. - A zero term limit is necessary for convergence but does not settle the series by itself.

1. \(\lim_{n\to\infty}a_n=0\). 2. A zero term limit is necessary for convergence but is not sufficient. 3. The \(n\)th-term test is inconclusive.

The \(n\)th-term test is inconclusive.

- Decide whether \(a_n\to0\) is a necessary condition, a sufficient condition, or both. - To disprove a universal statement, one counterexample is enough. - Use a familiar divergent series whose terms approach \(0\).

1. The statement is false. 2. The condition \(a_n\to0\) is necessary but not sufficient. 3. For example, \(\sum_{n=1}^{\infty}\frac1n\) diverges even though \(\frac1n\to0\).

False. For example, \(\sum_{n=1}^{\infty}\frac1n\) diverges even though its terms approach \(0\).

- Recall the necessary term behavior for every convergent series. - Compare the given limiting value with \(0\). - A nonzero term limit lets the \(n\)th-term test give a definitive conclusion.

1. The term limit is \(7\ne0\). 2. A convergent series must have terms approaching \(0\). 3. Therefore the statement is true.

True. Because \(\lim_{n\to\infty}a_n=7\ne0\), the \(n\)th-term test proves that the series diverges.

- Combine the two roots before finding the limit. - Rewrite the result as a constant raised to a variable exponent. - Check whether the limiting term is zero.

1. Simplify the term using properties of roots: \(a_n=\sqrt[n]{2}=2^{1/n}\). 2. Since \(2^{1/n}\to1\), the terms do not approach \(0\). 3. Therefore the series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(1\).

- Divide the numerator and denominator of the general term by \(n\). - Compare the resulting term limit with the necessary value \(0\) for series convergence. - A nonzero term limit proves divergence; a zero limit would only make this test inconclusive.

1. \(\lim_{n\to\infty}a_n=1\). 2. Because \(1\ne0\), the necessary condition for convergence fails. 3. The series diverges.

The series diverges because \(\lim_{n\to\infty}a_n=1\ne0\).

- Evaluate the sine expression for several consecutive integer values of \(n\). - Look for a repeating term pattern and ask whether it approaches \(0\). - Use only the necessary condition for convergence, as the prompt requires.

1. The term sequence \(a_n=\sin\left(\frac{\pi n}{2}\right)\) has no limit. 2. In particular, \(a_n\) does not approach \(0\). 3. Therefore the series diverges.

The series diverges by the \(n\)th-term test.

- Divide the rational term by the highest power of \(n\) in the denominator. - Determine whether the resulting limit equals \(0\). - If the term limit is nonzero, the series cannot converge, regardless of its starting index.

1. \(\lim_{n\to\infty}a_n=3\). 2. Because \(3\ne0\), the necessary condition for convergence fails. 3. The series diverges.

The series diverges because \(\lim_{n\to\infty}a_n=3\ne0\).

- Recall the horizontal asymptote of \(y=\arctan x\) as \(x\to\infty\). - Compare that term limit with the necessary value \(0\) for convergence of a series. - State only the conclusion justified by the \(n\)th-term test.

1. \(\lim_{n\to\infty}a_n=\frac{\pi}{2}\). 2. Because \(\frac{\pi}{2}\ne0\), the necessary condition for convergence fails. 3. The series diverges.

The series diverges because \(\lim_{n\to\infty}a_n=\frac{\pi}{2}\ne0\).

- Recognize the standard sequence \(\left(1+\frac1n\right)^n\). - Recall its limiting value as \(n\to\infty\). - A term limit different from \(0\) makes the necessary condition for convergence fail.

1. \(\lim_{n\to\infty}a_n=e\). 2. Because \(e\ne0\), the necessary condition for convergence fails. 3. The series diverges.

The series diverges because \(\lim_{n\to\infty}a_n=e\ne0\).

- The sign alternates, but the magnitude of the general term is \(\frac1n\). - Use the magnitude to determine the term limit. - If that limit is \(0\), this test alone cannot decide whether the series converges.

1. \(\lim_{n\to\infty}a_n=0\). 2. A zero term limit is necessary for convergence but is not sufficient. 3. The \(n\)th-term test is inconclusive.

The \(n\)th-term test is inconclusive.

- Compare the growth rates of \(\ln n\) and \(n\), or analyze \(\frac{\ln x}{x}\) as \(x\to\infty\). - Determine whether the summand approaches \(0\). - If it does, state that the \(n\)th-term test is inconclusive rather than claiming convergence.

1. \(\lim_{n\to\infty}a_n=0\). 2. A zero term limit is necessary for convergence but is not sufficient. 3. The \(n\)th-term test is inconclusive.

The summand approaches \(0\), so the \(n\)th-term test is inconclusive.

- Compare the leading coefficients of the quadratic numerator and denominator. - The \(n\)th-term test is inconclusive only when the resulting term limit equals \(0\). - Solve that condition for the parameter \(c\).

1. The term limit is \(\frac{c}{2}\). 2. The test is inconclusive exactly when this limit equals \(0\). 3. Thus \(c=0\).

\(c=0\).

- Rewrite \(e^{-cn}\) as \(\left(e^{-c}\right)^n\). - For \(c>0\), determine the size of the base \(e^{-c}\) relative to \(1\). - A term limit of \(0\) makes the \(n\)th-term test inconclusive; use no stronger test here.

1. For every \(c>0\), \(e^{-cn}\to0\). 2. A zero term limit does not establish convergence. 3. The \(n\)th-term test is inconclusive for every \(c>0\).

The \(n\)th-term test is inconclusive for every \(c>0\).

- Use the leading quadratic coefficients to express the term limit in terms of \(c\). - Set that expression equal to the required limit and solve for \(c\). - Then compare the specified nonzero term limit with the necessary condition for series convergence.

1. The limit is \(\frac{c}{2}\). 2. Set \(\frac{c}{2}=2\), giving \(c=4\). 3. The term limit is then \(2\ne0\), so the series diverges.

\(c=4\), and the series diverges.

- State the \(n\)th-term test as a one-way implication about nonzero or nonexistent term limits. - Identify what the test says—and does not say—when the term limit equals \(0\). - Separate the correctness of the computed limit from the validity of the student’s conclusion.

1. The calculation \(\lim_{n\to\infty}1/n=0\) is correct. 2. The test states that a nonzero or nonexistent term limit forces divergence. 3. A zero term limit gives no conclusion, so the student's reasoning is invalid.

The term limit is \(0\), so the \(n\)th-term test is inconclusive; it does not prove convergence.

- Look for a familiar divergent series whose terms nevertheless shrink toward \(0\). - A standard \(p\)-series at the boundary value provides such an example. - Verify both required properties: divergence of the series and a zero term limit.

1. The harmonic series \(\sum_{n=1}^{\infty}\frac1n\) diverges. 2. Its terms satisfy \(\lim_{n\to\infty}\frac1n=0\). 3. Therefore a divergent series can have terms approaching \(0\).

A counterexample is \(\sum_{n=1}^{\infty}\frac1n\), which diverges although \(1/n\to0\).

- Compute the term limit for each option independently. - The test proves divergence only for an option whose term limit is nonzero or does not exist. - Do not classify an option as convergent merely because its terms approach \(0\).

1. a) The term limit is \(\frac12\ne0\), so the series diverges. 2. b), c), and d) have term limit \(0\), so the test is inconclusive for them.

a) only.

- Determine the limit of the geometric correction \(\left(-\frac12\right)^n\). - Add that limiting behavior to the constant part of \(a_n\). - Use the \(n\)th-term test if the resulting term limit is not \(0\).

1. \(\left(-\frac12\right)^n\to0\), so \(a_n\to3\). 2. Because the term limit is nonzero, the series diverges.

The series diverges.

- A sequence limit concerns individual terms for large indices, not an average of positive and negative values. - Compare the even-indexed and odd-indexed subsequences of \((-1)^n\). - If those subsequences approach different values, the term limit does not exist.

1. The sequence \((-1)^n\) alternates between \(1\) and \(-1\). 2. A sequence limit is not an average or cancellation of values. 3. The limit does not exist, so \(\sum(-1)^n\) diverges by the \(n\)th-term test.

The limit does not exist, and the series diverges.

- First determine the limit of the inner expression \(\frac1n\). - Use continuity of the cosine function to pass that limit through \(\cos\). - A nonzero term limit proves divergence by the \(n\)th-term test.

1. As \(n\to\infty\), \(1/n\to0\). 2. By continuity, \(\cos(1/n)\to\cos0=1\). 3. The series diverges.

The series diverges.

- Rewrite the variable root using an exponential and a logarithm. - Analyze the exponent before returning to the original expression. - A zero term limit is necessary for convergence but is not sufficient.

1. Write \(n^{1/n}=e^{(\ln n)/n}\). 2. Since \(\frac{\ln n}{n}\to0\), \(n^{1/n}\to e^0=1\). 3. Therefore \(n^{1/n}-1\to0\). 4. The \(n\)th-term test is inconclusive; it does not prove either convergence or divergence.

The term limit is \(0\), so the \(n\)th-term test is inconclusive.

- Scale the radical and the other denominator term by the dominant power of the index. - Evaluate the simplified term limit. - Compare that limit with the necessary value for a convergent series.

1. Divide the numerator and denominator by \(n\): \(a_n=\frac{1}{\sqrt{1+1/n}+1}\). 2. Taking the limit gives \(a_n\to\frac{1}{1+1}=\frac12\). 3. Since the term limit is nonzero, the series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(\frac12\).

- Use the fixed bound on the trigonometric numerator. - Compare the magnitude of the term with a simpler expression that approaches zero. - Distinguish a necessary condition from a sufficient condition.

1. Since \(|\sin n|\le1\), \(\left|\frac{\sin n}{n^{1/3}}\right|\le\frac{1}{n^{1/3}}\). 2. The upper bound approaches \(0\), so the general term approaches \(0\) by the squeeze theorem. 3. Therefore the \(n\)th-term test is inconclusive; it does not establish convergence.

The term limit is \(0\), so the \(n\)th-term test is inconclusive.

- Rewrite each product factor as a quotient of consecutive integers. - Which factors cancel when the finite product is expanded? - Check the limiting value of the simplified term before considering partial sums.

1. Each factor satisfies \(1+1/k=(k+1)/k\). 2. The finite product cancels consecutively: \(\prod_{k=1}^{n}(k+1)/k=n+1\). 3. The series term is therefore \(a_n=(n+1)/n=1+1/n\to1\). 4. Since the terms do not approach \(0\), the series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(1\), not \(0\).

- Evaluate the finite numerator separately for even and odd indices. - Divide each possible numerator by the growing index. - State only the conclusion justified by a zero term limit.

1. The finite alternating sum \(\sum_{k=1}^{n}(-1)^k\) equals \(0\) when \(n\) is even and \(-1\) when \(n\) is odd. 2. Thus \(a_n\) equals \(0\) for even \(n\) and \(-\frac1n\) for odd \(n\). 3. In both cases, \(a_n\to0\). 4. Therefore the \(n\)th-term test is inconclusive.

The term limit is \(0\), so the \(n\)th-term test is inconclusive.

- Examine the terms along the infinitely many special indices. - A sequence limit must also be the limit of every subsequence. - Check the necessary term condition for convergence.

1. Along square indices, \(a_{m^2}=1\) for every positive integer \(m\). 2. Thus the term sequence has a subsequence that remains equal to \(1\). 3. Consequently \(a_n\) does not approach \(0\). 4. The series diverges by the \(n\)th-term test.

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

- Scale the integration variable by the moving upper limit. - Observe whether the transformed integral still depends on the index. - Compare the resulting term with the necessary zero limit.

1. Substitute \(u=x/n\), so \(dx=n\,du\) and the limits become \(0\) and \(1\). 2. Then \(a_n=\int_0^1 e^{-u}\,du=1-e^{-1}\). 3. Every term equals the same positive constant, so \(a_n\not\to0\). 4. The series diverges by the \(n\)th-term test.

The series diverges because every term equals \(1-e^{-1}\).

- Replace the finite arithmetic sum with a closed expression. - Cancel the dominant power of the index. - Apply the necessary term condition for convergence.

1. The numerator is \(\frac{n(n+1)}{2}\). 2. Thus \(a_n=\frac{n(n+1)}{2n^2}=\frac12\left(1+\frac1n\right)\). 3. Therefore \(a_n\to\frac12\ne0\). 4. The series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(\frac12\).

- Compare the long-term growth rates of a fixed power and an exponential. - Focus only on the general-term limit. - Do not infer convergence from a zero limit.

1. Exponential growth dominates any fixed real power, so \(\frac{n^p}{2^n}\to0\). 2. Therefore the necessary term condition for convergence is satisfied for every real \(p\). 3. The \(n\)th-term test is inconclusive for every \(p\); it does not prove convergence.

For every real \(p\), the term limit is \(0\), so the \(n\)th-term test is inconclusive.

- Separate the student's limit calculation from the conclusion drawn from it. - Recall which term-limit outcomes force divergence. - State only what a zero limit guarantees.

1. The term limit is indeed \(\lim_{n\to\infty}\frac{\ln n}{n}=0\). 2. The \(n\)th-term test proves divergence only when the term limit is nonzero or fails to exist. 3. A zero term limit is only a necessary condition for convergence. 4. Therefore the test is inconclusive and does not justify the student's convergence claim.

The student's conclusion is unsupported. The \(n\)th-term test is inconclusive because the terms approach \(0\).

- Begin with the necessary behavior of terms from the original convergent series. - Pass that limit through the continuous transformation. - Do not confuse a necessary condition with a convergence proof.

1. Convergence of \(\sum a_n\) implies \(a_n\to0\). 2. Continuity of sine gives \(\sin(a_n)\to\sin0=0\). 3. Thus the transformed terms satisfy the necessary zero-term condition. 4. The \(n\)th-term test is inconclusive; the information does not prove convergence of \(\sum\sin(a_n)\).

The transformed terms approach \(0\), so the \(n\)th-term test is inconclusive.

- Convert the squared-term information into information about magnitudes. - Is sign information needed to decide whether the terms approach zero? - Apply the necessary term condition.

1. Since \(a_n^2\to9\), \(|a_n|=\sqrt{a_n^2}\to3\). 2. Therefore the magnitudes of the terms do not approach \(0\). 3. Hence \(a_n\not\to0\), regardless of whether its signs stabilize or oscillate. 4. The series diverges by the \(n\)th-term test.

The series diverges because \(|a_n|\to3\), so \(a_n\not\to0\).

- Replace the repeated small expression with one variable. - Recognize a standard trigonometric limit. - State only what a zero term limit establishes.

1. Set \(u=1/\sqrt n\). Then \(u\to0\). 2. Since \(\frac{\tan u}{u}\to1\), the general term approaches \(1-1=0\). 3. The \(n\)th-term test is therefore inconclusive; it does not prove convergence.

The term limit is \(0\), so the \(n\)th-term test is inconclusive.

- Use the logarithm identity \(\ln(2n)=\ln n+\ln2\). - Divide the numerator and denominator by \(\ln n\) before taking the limit. - Compare the term limit with \(0\), the necessary value for convergence of a series.

1. \(\lim_{n\to\infty}a_n=1\). 2. Because \(1\ne0\), the necessary condition for convergence fails. 3. The series diverges.

The series diverges because \(\lim_{n\to\infty}a_n=1\ne0\).

- Let \(x=\frac1n\), so the term becomes a quotient involving \(\sin x\) and \(x\). - Use the standard limit \(\lim_{x\to0}\frac{\sin x}{x}\). - Compare the resulting term limit with \(0\) to classify the series by this test.

1. \(\lim_{n\to\infty}a_n=1\). 2. Because \(1\ne0\), the necessary condition for convergence fails. 3. The series diverges.

The series diverges because \(\lim_{n\to\infty}a_n=1\ne0\).

- Divide the numerator and denominator by \(n\) to express the term limit in terms of \(k\). - The test proves divergence exactly when that limit is not \(0\). - Handle separately the parameter value that makes the leading coefficient vanish.

1. \(\lim_{n\to\infty}\frac{kn+1}{n+2}=k\). 2. The test proves divergence when \(k\ne0\). 3. When \(k=0\), the term limit is \(0\), so this test is inconclusive.

The test proves divergence for \(k\ne0\); it is inconclusive for \(k=0\).

- Examine \(a_n\) separately along even and odd indices. - The reciprocal term vanishes, so determine what remains in each subsequence. - Identify the parameter value for which both subsequences can approach \(0\).

1. If \(c\ne0\), the term sequence oscillates between values approaching \(c\) and \(-c\), so it has no limit. 2. If \(c=0\), then \(a_n=1/n\to0\). 3. Therefore the test proves divergence for \(c\ne0\) and is inconclusive for \(c=0\).

The test proves divergence for \(c\ne0\); it is inconclusive for \(c=0\).

- Analyze the term sequence \(q^n\) according to whether \(|q|<1\), \(|q|=1\), or \(|q|>1\). - In the boundary cases, check \(q=1\) and \(q=-1\) separately. - The test proves divergence exactly when the terms fail to approach \(0\).

1. If \(|q|<1\), then \(q^n\to0\), so the test is inconclusive. 2. If \(q=1\), then \(q^n=1\), and if \(q=-1\), the terms oscillate. 3. If \(|q|>1\), the term magnitudes do not approach \(0\). 4. Thus the test proves divergence when \(|q|\ge1\).

The test proves divergence for \(|q|\ge1\) and is inconclusive for \(|q|<1\).

- Treat \(p>0\), \(p=0\), and \(p<0\) as separate cases. - For \(p<0\), rewrite \(n^p\) as a reciprocal power before taking the term limit. - Compare each case’s term limit with \(0\) and state whether the test proves divergence or is inconclusive.

1. If \(p>0\), the term limit is \(1\), so the series diverges. 2. If \(p=0\), every term is \(\frac12\), so the series diverges. 3. If \(p<0\), the term limit is \(0\), so the test is inconclusive.

For \(p\ge0\), the series diverges by the test. For \(p<0\), the test is inconclusive.

- Combine the two terms into a single fraction before taking the limit. - Check whether the coefficient of \(n\) cancels in the numerator. - Determine whether the remaining constant numerator divided by \(n+4\) approaches \(0\) for all or only some values of \(b\).

1. The general term simplifies to \(\frac{5n+b-5n-20}{n+4}=\frac{b-20}{n+4}\). 2. This term approaches \(0\) for every real \(b\). 3. Therefore the test is inconclusive for every real \(b\).

The test is inconclusive for every real \(b\).

- Write the necessary condition that every convergent series must satisfy. - Consider what its contrapositive allows you to prove. - Compare one convergent and one divergent series whose terms both approach \(0\).

1. A convergent series must have terms approaching \(0\). 2. Therefore a nonzero or nonexistent term limit guarantees divergence. 3. However, some series with terms approaching \(0\) converge and others diverge, so a zero limit cannot settle convergence.

It can prove divergence when the term limit is not \(0\), but a zero term limit is inconclusive.

- Divide the numerator and denominator by \(n\). - Bound \(\frac{(-1)^n}{n}\) in absolute value to determine its limit. - Compare the resulting term limit with \(0\) before stating the series conclusion.

1. Divide numerator and denominator by \(n\): \(a_n=\frac{1}{1+\frac{(-1)^n}{n}}\). 2. Since \(\frac{(-1)^n}{n}\to0\), \(a_n\to1\). 3. The series diverges.

The series diverges.

- Let \(x=\frac1n\) so that \(x\to0^+\). - Rewrite the term as a quotient involving \(e^x-1\). - What standard limit or derivative definition evaluates that quotient?

1. Let \(x=\frac1n\), so \(x\to0^+\). 2. \(n\left(e^{1/n}-1\right)=\frac{e^x-1}{x}\to1\). 3. The term limit is nonzero, so the series diverges.

The series diverges.

- Direct substitution gives an indeterminate difference; what conjugate can rationalize it? - After rationalizing, divide by the highest power of \(n\) present. - Does the resulting term limit equal \(0\)?

1. Rationalize: \(\sqrt{n^2+4n}-n=\frac{4n}{\sqrt{n^2+4n}+n}\). 2. Divide numerator and denominator by \(n\) to get a limit of \(\frac{4}{\sqrt{1+\frac4n}+1}=2\). 3. The series diverges.

The series diverges.

- Recall a familiar convergent series whose terms approach \(0\). - Recall a familiar divergent series whose terms also approach \(0\). - What does the contrast show about the converse of the necessary condition?

1. \(\sum_{n=1}^{\infty}\left(\frac12\right)^n\) converges and its terms approach \(0\). 2. \(\sum_{n=1}^{\infty}\frac1n\) diverges and its terms also approach \(0\). 3. Therefore a zero term limit does not determine convergence.

One example pair is \(\sum_{n=1}^{\infty}\left(\frac12\right)^n\) and \(\sum_{n=1}^{\infty}\frac1n\); the test is inconclusive when the term limit is \(0\).

- Rewrite the expression so the same small quantity appears inside the logarithm and in a denominator. - Compare with a standard logarithmic limit near zero. - A convergent series cannot have terms approaching a nonzero number.

1. Let \(u=\frac{3}{n}\). Then \(u\to0\), and \(\frac{\ln(1+3/n)}{1/n}=3\frac{\ln(1+u)}{u}\). 2. The standard logarithmic limit gives \(\frac{\ln(1+u)}{u}\to1\). 3. Therefore the general term approaches \(3\ne0\). 4. By the \(n\)th-term test, the series diverges.

The series diverges because its terms approach \(3\), not \(0\).

- Introduce a variable for the small angle. - Rewrite the outside power of the index in terms of that angle. - Determine whether the resulting term limit is zero.

1. Set \(u=\frac{2}{n}\). Then \(n^2\left(1-\cos\left(\frac{2}{n}\right)\right)=4\frac{1-\cos u}{u^2}\). 2. As \(u\to0\), \(\frac{1-\cos u}{u^2}\to\frac12\). 3. Thus the general term approaches \(4\cdot\frac12=2\ne0\). 4. The series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(2\).

- Evaluate the definite integral before taking the sequence limit. - Rewrite the logarithmic difference as the logarithm of a ratio. - Check the necessary term condition for convergence.

1. Evaluate the integral: \(a_n=n\ln(n+x)\big|_0^1=n\ln\left(1+\frac{1}{n}\right)\). 2. The standard exponential-logarithmic limit gives \(n\ln\left(1+\frac{1}{n}\right)\to1\). 3. Since \(a_n\to1\ne0\), the series diverges by the \(n\)th-term test.

The series diverges because \(a_n\to1\).

- Separate the given term into the nearby reference quantity and an error. - Determine what the error bound forces as \(n\) grows. - Apply the necessary condition for a convergent series only after finding the term limit.

1. Since \(1/n\to0\), \(\cos(1/n)\to1\). 2. The bound gives \(a_n-\cos(1/n)\to0\). 3. Therefore \(a_n=\cos(1/n)+[a_n-\cos(1/n)]\to1\). 4. Because the terms do not approach \(0\), the series diverges by the \(n\)th-term test.

The series diverges because \(a_n\to1\ne0\).

- Relate the inverse tangent of a positive number to that of its reciprocal. - Rewrite the expression using a small input approaching zero. - Compare with the standard inverse-tangent limit near zero.

1. For \(n>0\), \(\arctan n+\arctan(1/n)=\frac{\pi}{2}\). 2. Hence \(n\left(\arctan n-\frac{\pi}{2}\right)=-n\arctan(1/n)\). 3. Since \(\frac{\arctan u}{u}\to1\) as \(u\to0\), the term limit is \(-1\). 4. The series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(-1\).

- Translate convergence to zero into an eventual inequality involving an arbitrary tolerance. - Choose a tolerance smaller than the fixed lower bound. - Use the infinitely many exceptional indices to contradict eventual smallness.

1. If \(a_n\to0\), then for \(\varepsilon=c/2\) there would be an index \(N\) such that \(|a_n|<c/2\) for every \(n\ge N\). 2. The hypothesis gives an index \(n\ge N\) with \(|a_n|\ge c\), contradicting that inequality. 3. Therefore \(a_n\not\to0\). 4. The series diverges by the \(n\)th-term test.

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

- Rewrite the fraction as one plus a quantity approaching zero. - Account for the small shift between the denominator and the exponent. - Identify the resulting exponential-type limit.

1. Rewrite \(\left(\frac{n-1}{n+1}\right)^n=\left(1-\frac{2}{n+1}\right)^n\). 2. Taking logarithms or using the standard exponential limit gives \(\left(1-\frac{2}{n+1}\right)^n\to e^{-2}\). 3. Since \(e^{-2}\ne0\), the series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(e^{-2}\).

- Introduce one variable for the common small input. - Insert that input as a numerator and denominator to separate the expression. - Evaluate the two standard limits before applying the test.

1. Set \(u=\frac1n\). Then \(u\to0\), and \(\frac{e^{1/n}-1}{\sin(1/n)}=\frac{e^u-1}{u}\cdot\frac{u}{\sin u}\). 2. Both factors approach \(1\), so the general term approaches \(1\). 3. Since the term limit is nonzero, the series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(1\).

- Treat the two given expressions as a system involving the original sequences. - Combine them to isolate each term sequence. - Compare the resulting limits with the necessary term condition.

1. Add the two limiting relations: \(2a_n=(a_n+b_n)+(a_n-b_n)\to6\), so \(a_n\to3\). 2. Subtract the second relation from the first: \(2b_n=(a_n+b_n)-(a_n-b_n)\to-6\), so \(b_n\to-3\). 3. Neither term sequence approaches \(0\). 4. Both series diverge by the \(n\)th-term test.

Both series diverge: \(a_n\to3\) and \(b_n\to-3\).

- Factor the dominant exponential term before applying the logarithm rule. - Compare the remaining correction with zero. - Divide the final numerator and denominator by the index.

1. Factor \(e^n\) inside the logarithm: \(\ln(e^n+n)=n+\ln(1+ne^{-n})\). 2. Since \(ne^{-n}\to0\), \(\ln(1+ne^{-n})\to0\). 3. Therefore \(a_n=\frac{n}{n+\ln(1+ne^{-n})}\to1\). 4. The series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(1\).

- Identify the standard limit hidden in the variable power. - Set the resulting term limit equal to the value required for possible convergence. - Distinguish what the test proves from what it cannot decide.

1. The standard exponential limit gives \(\left(1+\frac{c}{n}\right)^n\to e^c\). 2. Therefore the series terms approach \(e^c-2\). 3. This limit equals \(0\) exactly when \(e^c=2\), so \(c=\ln2\). 4. For \(c\ne\ln2\), the term limit is nonzero and the series diverges by the \(n\)th-term test. 5. For \(c=\ln2\), the terms approach \(0\), so the test is inconclusive.

The \(n\)th-term test is inconclusive only for \(c=\ln2\). For every \(c\ne\ln2\), it proves that the series diverges.

- Bound the greatest-integer expression between two nearby real numbers. - Divide the entire inequality by the positive index. - Identify when the resulting limit fails the necessary zero condition.

1. The floor inequality gives \(n\alpha-1<\lfloor n\alpha\rfloor\le n\alpha\). 2. Dividing by \(n>0\) gives \(\alpha-\frac1n<a_n\le\alpha\). 3. Therefore \(a_n\to\alpha\). 4. If \(\alpha\ne0\), the term limit is nonzero and the series diverges by the \(n\)th-term test. If \(\alpha=0\), the test does not prove divergence.

The \(n\)th-term test proves divergence exactly when \(\alpha\ne0\).

- Assume the term sequence had the limit required for convergence. - What would then happen to the shifted sequence? - Compare the resulting difference limit with the given one.

1. Suppose for contradiction that \(a_n\to0\). 2. Then the shifted sequence also satisfies \(a_{n+1}\to0\). 3. Limit laws would give \(a_{n+1}-a_n\to0-0=0\), contradicting the given limit \(4\). 4. Therefore \(a_n\not\to0\), and the series diverges by the \(n\)th-term test.

The series diverges because \(a_n\) cannot approach \(0\).

- Look for a familiar positive series at the boundary of convergence. - Verify the term limit separately from the behavior of partial sums. - Compare each finite partial sum with an improper-integral expression that grows without bound.

1. Choose the harmonic series \(\sum_{n=1}^{\infty}1/n\). 2. Its terms satisfy \(1/n\to0\). 3. For every positive integer \(N\), \(\sum_{n=1}^{N}\frac1n\ge\int_1^{N+1}\frac{1}{x}\,dx=\ln(N+1)\). 4. Since \(\ln(N+1)\to\infty\), the partial sums diverge to \(+\infty\). Thus a zero term limit does not imply series convergence.

The harmonic series is an example: \(1/n\to0\), but \(\sum1/n\) diverges.

- If the terms approach a limit \(L\), what fixed-point equation must \(L\) satisfy? - What happens when you subtract that candidate limit from the recurrence? - Does the resulting term limit satisfy the necessary condition for series convergence?

1. A limiting term value \(L\) must satisfy \(L=\frac12L+\frac32\), so \(L=3\). 2. Subtracting \(3\) from the recurrence gives \(a_{n+1}-3=\frac12(a_n-3)\). 3. Thus \(a_n-3=-3\left(\frac12\right)^{n-1}\to0\), so \(a_n\to3\). 4. Because the terms do not approach \(0\), the series diverges.

The series diverges.

- Separate the general term according to the parity of the index. - Recognize a standard exponential limit in each case. - Compare the two subsequential limits before applying the test.

1. Along even indices, \(a_{2m}=\left(1+\frac{1}{2m}\right)^{2m}\to e\). 2. Along odd indices, \(a_{2m+1}=\left(1-\frac{1}{2m+1}\right)^{2m+1}\to e^{-1}\). 3. The two subsequences have different limits, so \((a_n)\) has no limit and in particular does not approach \(0\). 4. The series diverges by the \(n\)th-term test.

The series diverges because its terms have subsequential limits \(e\) and \(e^{-1}\), rather than approaching \(0\).

- Rewrite the variable root using an exponential. - Match the difference of exponentials to a standard limit near zero. - Determine when the resulting limiting constant is nonzero.

1. Write \(c^{1/n}=e^{(\ln c)/n}\). 2. Using \(\lim_{u\to0}\frac{e^u-1}{u}=1\), \(n\left(c^{1/n}-1\right)=\ln c\cdot\frac{e^{(\ln c)/n}-1}{(\ln c)/n}\to\ln c\) when \(c\ne1\). 3. If \(c\ne1\), then \(\ln c\ne0\), so the series diverges by the \(n\)th-term test. 4. If \(c=1\), every term is \(0\); the test's zero-limit condition does not prove divergence.

The \(n\)th-term test proves divergence for every \(c>0\) with \(c\ne1\). At \(c=1\), the terms are all \(0\).

- Rewrite the given expression to measure how far the positive term is from an equality case. - Positivity lets you introduce a square root without changing the essential condition. - Decide what the original terms must approach before applying the necessary condition for convergence.

1. Let \(x_n=\sqrt{a_n}>0\). Then \(\left(x_n-\frac{1}{x_n}\right)^2=a_n+\frac{1}{a_n}-2\to0\). 2. Therefore \(x_n-\frac{1}{x_n}\to0\). 3. Also \(\left(x_n+\frac{1}{x_n}\right)^2=\left(x_n-\frac{1}{x_n}\right)^2+4\to4\). Because the expression is positive, \(x_n+\frac{1}{x_n}\to2\). 4. Adding the two limits gives \(2x_n\to2\), so \(x_n\to1\) and \(a_n\to1\). 5. Since the terms do not approach \(0\), \(\sum a_n\) diverges by the \(n\)th-term test.

The series must diverge because \(a_n\to1\ne0\).

- Rewrite the base using one plus a small quantity. - Take a logarithm to bring the variable exponent down as a factor. - Compare the growth of the square root with the reciprocal-size logarithm.

1. Let \(a_n=\left(1+\frac1n\right)^{-\sqrt n}\). 2. Its logarithm is \(\ln a_n=-\sqrt n\ln\left(1+\frac1n\right)\). 3. Rewrite this as \(\ln a_n=-\frac{1}{\sqrt n}\cdot\frac{\ln(1+1/n)}{1/n}\). The first factor approaches \(0\), and the second approaches \(1\), so \(\ln a_n\to0\). 4. Therefore \(a_n\to e^0=1\), and the series diverges by the \(n\)th-term test.

The series diverges because its terms approach \(1\).

- How close is the square root to the nearby integer-sized expression? - Can rewriting the small difference make its limit easier to see? - How does the periodic behavior of cosine respond to the integer-sized part?

1. Write \(\sqrt{n^2+n}=n+d_n\), where \(d_n=\sqrt{n^2+n}-n=\frac{1}{\sqrt{1+1/n}+1}\to\frac12\). 2. Then \(a_n=\cos(\pi n+\pi d_n)=(-1)^n\cos(\pi d_n)\). 3. Since \(d_n\to\frac12\), \(\cos(\pi d_n)\to0\). Thus \(|a_n|\to0\), so \(a_n\to0\). 4. The \(n\)th-term test is inconclusive.

The term limit is \(0\), so the \(n\)th-term test is inconclusive.