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True or False The series \(\sum\limits_{k\,=\,1}^{\infty }(-1)^{k}\cos ( k\pi )\) is an alternating series.

Question

True or False \(\sum\limits_{k\,=\,1}^{\infty }[ 1+(-1) ^{k}]\) is an alternating series.

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True or False In an alternating series, \(\sum\limits_{k\,=\,1}^{\infty }(-1)^{k}\,a_{k}\), if \(\lim\limits_{n\, \rightarrow \,\infty }\,a_{n}=0,\) then the series is convergent.

589

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True or False If the alternating series \(\sum\limits_{k\,=\,1}^{\infty }(-1)^{k+1}\,a_{k}\) satisfies the two conditions of the Alternating Series Test, then the error \(E_{n}\) in using the sum \(S_{n}\) of the first \(n\) terms as an approximation of the sum \(S\) of the series is \(\vert E_{n}\vert \leq a_{n}\).

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True or False A series that is not absolutely convergent is divergent.

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True or False If a series is absolutely convergent, then the series converges.

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

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

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

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

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

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

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

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

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

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\(\sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{1}{k!}\)

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

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

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

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\(\sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{1}{k!}\)

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

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

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\(\sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{1}{k!}\left( \dfrac{1}{3}\right) ^{k}\)

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\(\sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{1}{k!}\left( \dfrac{1}{2}\right) ^{k}\)

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\(\sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{1}{2k+1}\left( \dfrac{1}{3}\right) ^{2k+1}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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\(1-\dfrac{1}{2!}+\dfrac{1}{3!}-\dfrac{1}{4!}+\dfrac{1}{5!}-\cdots\)

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\(1-\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}-\dfrac{1}{7^{2}}+\cdots\)

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Show that the positive terms of \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{(-1)^{k+1}}{k}\) diverge.

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Show that the negative terms of \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{(-1)^{k+1}}{k}\) diverge.

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Show that the terms of the series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{(-1)^{k+1}}{k}\) can be rearranged so the resulting series converges to \(0\).

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Show that the terms of the series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{(-1)^{k+1}}{k}\) can be rearranged so the resulting series converges to \(2\).

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Show that the terms of the series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{(-1)^{k+1}}{k}\) can be rearranged so the resulting series diverges.

Question

Show that the series \[ e^{-x}\cos x+e^{-2x}\cos (2x) +e^{-3x}\cos (3x)+\cdots \]

is absolutely convergent for all positive values of \(x\). (Hint: Use the fact that \(\vert \cos \theta \vert \leq 1\).)

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Determine whether the series below converges (absolutely or conditionally) or diverges. \[ 1+r\cos \theta +r^{2}\cos\! \left( 2\theta \right) +r^{3}\cos (3\theta) +\cdots \]

Question

What is wrong with the following argument? \begin{eqnarray*} A &=& 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{7}-\dfrac{1}{8}+\cdots \\[3pt] \left(\dfrac{1}{2}\right) A &=& \dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{8}+\cdots \end{eqnarray*} So, \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A+\left(\dfrac{1}{2}\right) A = 1+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{4}+\cdots \]

The series on the right is a rearrangement of the terms of the series \(A\). So its sum is \(A\), meaning \[ \begin{array}{l} A + \left( {\frac{1}{2}} \right)A &=& A \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A &=& 0 \\ \end{array} \]

But, \[ A = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \left( {\frac{1}{5} - \frac{1}{6}} \right) + \cdots > 0 \]

Bernoulli’s Error In Problems 59–61, consider an incorrect argument given by Jakob Bernoulli to prove that \[ \sum\limits_{k = 1}^\infty {\frac{1}{{k(k + 1)}} = \frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + \cdots = 1} \]

Bernoulli’s argument went as follows: \[ Let \,N = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots ., Then\] \[N - 1 =\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots .\]

Now, subtract term-by-term to get \[\begin{array}{l} N - (N - 1) = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \cdots \,\,\,\,\,(4) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 = \frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + \cdots \\ \end{array}\]

Question

What is wrong with Bernoulli’s argument?

Question

In general, what can be said about the convergence or divergence of a series formed by taking the term-by-term difference (or sum) of two divergent series? Support your answer with examples.

Question

Although the method is wrong. Bernoulli’s conclusion is correct; that is, it is true that \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k(k+1) }=1.\) Prove it! [Hint: Look at the partial sums using the form of the series in (4).]

Question

Show that if two series \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) and \(\sum\limits_{k\,=\,1}^{\infty }b_{k}\) are absolutely convergent, the series \(\sum\limits_{k\,=\,1}^{\infty }(a_{k}+b_{k})\) is also absolutely convergent. Moreover, \(\sum\limits_{k\,=\,1}^{\infty}\,(a_{k}+b_{k})=\sum\limits_{k\,=\,1}^{\infty}a_{k}+\sum\limits_{k\,=\,1}^{\infty }b_{k}\).

Question

Show that if a series converges absolutely, then the series consisting of just the positive terms converges, as does the series consisting of just the negative terms.

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Show that if a series converges conditionally, then the series consisting of just the positive terms diverges, as does the series consisting of just the negative terms.

Question

Determine whether the series \(\sum\limits_{k=1}^{\infty }c_{k}\), where \[ \,\,{c_k} = \left\{ \begin{array}{l} \frac{1}{{{a^k}}}\,\,\,\,\qquad{\rm{if}}\,k\,{\rm{is}}\,{\rm{even}} \\ - \frac{1}{{{b^k}}}\qquad{\rm{if}}\,k\,{\rm{is}}\,{\rm{odd}} \\ \end{array} \right.a > 1,\,b > 1, \]

converges absolutely, converges conditionally, or diverges.

Question

Prove that if a series \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) is absolutely convergent, any rearrangement of its terms results in a series that is also absolutely convergent. (Hint: Use the triangle inequality: \(\left\vert a+b\right\vert \leq \vert a \vert+\left\vert b\right\vert\).)

Question

Suppose that the terms of the series \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) are alternating, \(\vert a_{k+1}\vert \leq \left\vert a_{k}\right\vert\) for all integers \(k,\) and \(\lim\limits_{n\rightarrow \infty }a_{n}=0.\) Show that the series converges.

Question

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

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\(\sum\limits_{k=2}^{\infty }\dfrac{(-1)^{k}}{\sqrt[p]{k^{3}+1}},~p>2\)

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

Question

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

Question

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

Question

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

Question

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

Question

Let \(\{a_{n}\}\) be a sequence that is decreasing and is bounded from below by \(0\). Define \[ R_{n}=\sum_{k=n+1}^{\infty }(-1)^{k+1}a_{k} \qquad \hbox{and}\qquad \Delta a_{k}=a_{k}-a_{k+1} \]

Suppose that the sequence \(\left\{ \Delta a_{k}\right\}\) decreases.

1. Show that the series \(\sum\limits_{k\,=\,1}^{\infty }(-1)^{k+1}~\Delta a_{k}\) is a convergent alternating series.
2. Show that \[ \left\vert R_{n}\right\vert =\frac{a_{n}}{2}+\frac{1}{2}\sum_{k\,=\,1}^{\infty }(-1)^{k}~\Delta \,a_{k+n-1} \] and \(\sum\limits_{k\,=\,1}^{\infty }(-1)^{k}~\Delta \,a_{k+n-1}<0\). Deduce \(\left\vert R_{n}\right\vert <\dfrac{1}{2}\,a_{n}\).
3. Show that \[ \left\vert R_{n}\right\vert =\frac{a_{n+1}}{2}+\frac{1}{2} \sum_{k\,=\,1}^{\infty }(-1)^{k+1}~\Delta \,a_{k+n} \]
4. Conclude that \(\dfrac{a_{n+1}}{2}<\left\vert R_{n}\right\vert\).

Source: Based on R. Johnsonbaugh, Summing an alternating series, American Mathematical Monthly, Vol. 86, No. 8 (Oct. 1979), pp. 637-648.