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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].

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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 \(\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}=0.\)

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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.

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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 \(\dfrac{1}{2}.\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1. 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. 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. 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.

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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.

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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.

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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.

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

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

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

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

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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.

Question

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

Question

1. Show that \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{\ln k}{k^{p}}\) diverges for \(0<p\leq 1.\)
2. 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.

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

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

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

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

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

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

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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.

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

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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 \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{ a}_{n}}{{ b}_{n}}\) \(=0.\)

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

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

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

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

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