8.4 Assess Your Understanding

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Concepts and Vocabulary

  1. Multiple Choice If each term of a series \(\sum\limits_{k=1}^{\infty }a_{k}\) of positive terms is greater than or equal to the corresponding term of a known divergent series \(\sum\limits_{k=1}^{\infty }c_{k}\) of positive terms, then the series \(\sum\limits_{k=1}^{\infty}a_{k}\) is [(a) convergent, (b) divergent].

(b)

  1. True or False Suppose \(\sum\limits_{k=1}^{\infty}\,a_{k}\) and \(\sum\limits_{k=1}^{\infty }\,b_{k}\) are both series of positive terms. The series \(\sum\limits_{k=1}^{\infty }\,a_{k}\) and the series \(\sum\limits_{k=1}^{\infty }\,b_{k}\) both converge or both diverge if \(\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}=0.\)

False

  1. True or False If \(\sum\limits_{k=1}^{\infty}\,a_{k}\) and \(\sum\limits_{k=1}^{\infty }\,b_{k}\) are both series of positive terms and if \(\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}= L,\) where \(L\) is a positive real number, then the series to be tested converges.

False

  1. True or False Since the \(p\)-series \(\sum\limits_{k=1}^{\infty}\dfrac{1}{k^{3/2}}\) converges and \(\lim\limits_{n\rightarrow \infty }\dfrac{\dfrac{1}{2n^{3/2}+5}}{\dfrac{1}{n^{3/2}}}=\dfrac{1}{2},\) then by the Limit Comparison Test, the series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{2k^{3/2}+5}\) converges to \(\dfrac{1}{2}.\)

False

Skill Building

In Problems 5–14, use the Comparison Tests for Convergence or Divergence to determine whether each series converges or diverges.

  1. \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k(k+1)}\): by comparing it with \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{2}}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{(k+2)^{2}}\): by comparing it with \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{2}}\)

  1. \(\sum\limits_{k=2}^{\infty }\dfrac{4^{k}}{7^{k}+1}\): by comparing it with \(\sum\limits_{k=2}^{\infty }\left(\dfrac{4}{7}\right)^{k}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty} \dfrac{1}{(2k-1)(2^{k})}\): by comparing it with \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2^{k}}\)

  1. \(\sum\limits_{k=2}^{\infty }\dfrac{1}{\sqrt{k(k-1)}}\): by comparing it with \(\sum\limits_{k=2}^{\infty }\dfrac{1}{k}\)

Diverges

  1. \(\sum\limits_{k=2}^{\infty }\dfrac{\sqrt{k}}{k-1}\): by comparing it with \(\sum\limits_{k=2}^{\infty }\dfrac{1}{\sqrt{k}}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k(k+1)(k+2)}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{6}{5k-2}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{\sin ^{2}k}{k^{\pi}}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{\cos ^{2}k}{k^{2}+1}\)

In Problems 15–28, use the Limit Comparison Test to determine whether each series converges or diverges.

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{(k+1)(k+2)}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{2}+1}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{\sqrt{k^{2}+1}}\)

Diverges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{\sqrt{k}}{k+4}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{3\sqrt{k}+2}{2k^{2}+5}\)

Converges

  1. \(\sum\limits_{k\,=\,2}^{\infty}\dfrac{3\sqrt{k}+2}{2k-3}\)

  1. \(\sum\limits_{k\,\,=\,2}^{\infty}\dfrac{1}{k\sqrt{k^{2}-1}}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{k}{(2k-1)^{2}}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{3k+4}{k2^{k}}\)

Converges

  1. \(\sum\limits_{k\,\,=\,2}^{\infty}\dfrac{k-1}{k2^{k}}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{2^{k}+1}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{5}{3^{k}+2}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{k+5}{k^{k+1}}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{5}{k^{k}+1}\)

In Problems 29–40, use any of the comparison tests to determine whether each series converges or diverges.

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{6k}{5k^{2}+2}\)

Diverges

  1. \(\sum\limits_{k\,=\,2}^{\infty}\dfrac{6k+3}{2k^{3}-2}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{7+k}{(1+k^{2})^{4}}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\left( \dfrac{7+k}{1+k^{2}}\right) ^{4}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{e^{1/k}}{k}\)

Diverges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{1+e^{k}}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{\left( 1+\dfrac{1}{k}\right) ^{2}}{e^{k}}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{k\,2^{k}}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1+\sqrt{k}}{k}\)

Diverges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\left(\dfrac{1+3\sqrt{k}}{k^{2}}\right)\)

  1. \(\sum\limits_{k\,=\,1}^{\infty}\left( \dfrac{1}{2}\right)^{k}\sin ^{2}k\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{\tan ^{-1}k}{k^{3}}\)

Applications and Extensions

In Problems 41–48, determine whether each series converges or diverges.

  1. \(\sum\limits_{k\,=\,2}^{\infty }\dfrac{2}{k^{3}\ln k}\)

Converges

  1. \(\sum\limits_{k=2}^{\infty }\dfrac{1}{\sqrt{k}(\ln k)^{4}}\)

  1. \(\sum\limits_{k\,=~2}^{\infty }\dfrac{\ln k}{k+3}\)

Diverges

  1. \(\sum\limits_{k\,=\,2}^{\infty }\dfrac{(\ln k)^{2}}{k^{5/2}}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty }\sin \dfrac{1}{k}\)

Diverges

  1. \(\sum\limits_{k\,=\,1}^{\infty }\tan \dfrac{1}{k}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k!}\)

Converges

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{k!}{k^{k}}\)

581

  1. It is known that \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{2}}\) is a convergent \(p\)-series.

    1. (a) Use the Comparison Test for Convergence with \(\sum\limits_{k=1}^{\infty } \dfrac{1}{k^{2}}\) to show that \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{2}+1}\) converges.
    2. (b) Explain why the Comparison Test for Convergence and \(\sum\limits_{k=2}^{\infty }\dfrac{1}{k^{2}}\) cannot be used to show \(\sum\limits_{k=2}^{\infty }\dfrac{1}{k^{2}-1}\) converges.
    3. (c) Can the Limit Comparison Test be used to show \(\sum\limits_{k\,=\,2}^{\infty }\dfrac{1}{k^{2}-1}\) converges? If it cannot, explain why.

See Student Solutions Manual.

  1. Show that any series of the form \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{d_{k}}{10^{k}}\), where the \(d_{k}\) are digits \((0, 1, 2, \ldots , 9)\), converges.

  1. Suppose the series \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) of positive terms is to be tested for convergence or divergence, and the series \(\sum\limits_{k\,=\,1}^{\infty }d_{k}\) of positive terms diverges. Show that if \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{a_{n}}{d_{n}}=\infty\), then \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) diverges.

See Student Solutions Manual.

  1. Suppose the series \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) of positive terms is to be tested for convergence or divergence, and the series \(\sum\limits_{k\,=\,1}^{\infty }d_{k}\) of positive terms converges. Show that if \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{a_{n}}{d_{n}}=0\), then \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) converges.

In Problems 53–56, use the results of Problems 51 and 52 to determine whether each of the following series converges or diverges.

  1. \(\sum\limits_{k\,=\,2}^{\infty }\dfrac{1}{\ln k}\)

Diverges

  1. \(\sum\limits_{k\,=\,2}^{\infty }\left( \dfrac{1}{\ln k}\right)^{2}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{\ln k}{k^{2}}\)

Converges

  1. \(\sum\limits_{k\,=\,2}^{\infty }\dfrac{1}{(k\ln k) ^{2}}\)

  1. Explain why the Limit Comparison Test and the series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{e^{k}},\) which converges, cannot be used to determine if the series \(\sum\limits_{k\,=2}^{\infty }\dfrac{1}{\ln k}\) converges or diverges.

Answers will vary.

    1. (a) Show that \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{\ln k}{k^{p}}\) converges for \(p>1\).
    2. (b) Show that \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{(\ln k) ^{r}}{k^{p}},\) where \(r\) is a positive number, converges for \(p>1\).
    1. (a) Show that \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{\ln k}{k^{p}}\) diverges for \(0<p\leq 1.\)
    2. (b) Show that \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{(\ln k) ^{r}}{k^{p}},\) where \(r\) is a positive real number and \(0<p\leq 1\), diverges.

  1. (a) See Student Solutions Manual.
  2. (b) See Student Solutions Manual.

In Problems 60–63, use the results of Problems 58 and 59 to determine whether each of the following series converges or diverges.

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{\ln k}{k}\)

  1. \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{\sqrt{\ln k}}{\sqrt{k}}\)

Diverges

  1. \(\sum\limits_{k\,=\,2}^{\infty }\dfrac{\ln k}{k^{3}}\)

  1. \(\sum\limits_{k\,=\,2}^{\infty }\dfrac{(\ln k)^{2}}{\sqrt{k^{3}}}\)

Converges

  1. Use the Comparison Tests for Convergence or Divergence to show that the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{p}}\):

    1. (a) converges if \(p>1\).
    2. (b) diverges if \(0<p\leq 1.\)

  1. If the series \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) of positive terms converges, show that the series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{a_{k}}{k}\) also converges.

See Student Solutions Manual.

  1. Show that \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{1+2^{k}}\) converges.

  1. It is known that the harmonic series \(\sum\limits_{k\,=\,1}^{\infty }b_{k}=\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\) diverges. Find two different series \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\), one that converges and one that diverges, so that \(\lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}}=0.\) These two examples show that the Limit Comparison Test gives no guidance if \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{ a}_{n}}{{ b}_{n}}\) \(=0.\)

Answers will vary; for example: \(\sum{\dfrac{1}{(k+1)\ln k}}\) diverges; \(\sum{\dfrac{1}{k^2}}\) converges

Challenge Problems

In Problems 68 and 69, determine whether each series converges or diverges.

  1. \(\sum\limits_{k\,=\,2}^{\infty }\dfrac{\ln (2k+1)}{\sqrt{k^{2}-2}\sqrt{k^{3}-2k-3}}\)

  1. \(\sum\limits_{k=1}^{\infty }\dfrac{\sqrt{k}}{\sqrt{( k^{3}-k+1) \ln (2k+1)}}\)

Diverges

  1. Show that the series \(\sum\limits_{k=1}^{\infty }\dfrac{1+\sin k}{4^{k}}\) converges.

  1. Show that the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{1+1/k}}\) diverges.

See Student Solutions Manual.

    1. (a) Determine whether the series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{k^{2}-3k-2}{k^{2}(k+1)^{2}}\) converges or diverges.
    2. (b) If it converges, find its sum.