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Show that the series converges, but not absolutely. $\sum_{n=1}^{\infty}( $exp$(\frac{(-1)^n}{n})-1)$.

My Try:

Let $a_n=$exp$(\frac{(-1)^n}{n})-1$. I was going to use alternating series test because the sequence $\{a_n\}$ is alternating. But $\{|a_n|\}$ is not decreasing to $0$. So, I am stuck now. Can anybody please give me a hint?

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    $\begingroup$ Can you give a useful estimate for $\lvert e^x - 1 - x\rvert$? $\endgroup$ – Daniel Fischer Jul 19 '15 at 20:34
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    $\begingroup$ Hint: Taylor's theorem shows that $e^t=1+t+E(t)$ where $|E(t)|\le cx^2$ for $|x|\la 1$. $\endgroup$ – David C. Ullrich Jul 19 '15 at 20:37
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    $\begingroup$ If $|a_n|$ doesn't decrease to 0, then the series can't converge, absolutely or conditionally. $\endgroup$ – Teepeemm Jul 20 '15 at 0:19
  • $\begingroup$ "But $|a_n|$ is not decreasing to $0$ ": are you sure ? $\endgroup$ – Yves Daoust Jul 20 '15 at 6:19
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    $\begingroup$ Possible duplicate of Convergence of $ \sum_{n=1}^\infty\left(\exp\left(\frac{(-1)^n}{n}\right)-1\right)$? $\endgroup$ – user557902851 Aug 15 '18 at 0:07
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You may just use the fact that, as $x \to 0$, using the Taylor expansion, $$ e^x=1+x+O(x^2) $$ giving, for some $n_0\geq1$, $$ \sum_{n\geq n_0}\left(\exp \left(\frac{(-1)^n}{n}\right)-1\right)=\sum_{n\geq n_0}\frac{(-1)^n}{n}+\sum_{n\geq n_0}O\left(\frac1{n^2}\right). $$ The latter series is absolutely convergent and the series $\displaystyle \sum_{n\geq n_0}\frac{(-1)^n}{n}$ is conditionally convergent. It gives the desired result.

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PART 1: ESTABLISHING CONVERGENCE

From the Mean Value Theorem we have

$$e^{(-1)^n/n}-1=\frac{(-1)^n}{n}+e^{\xi_n}\frac{1}{n^2}$$

for $0<|\xi_n|<\frac{1}{n}$. Then, we have

$$\begin{align} \sum_{n=1}^{\infty}\left(e^{(-1)^n/n}-1\right)&=\sum_{n=1}^{\infty}\left(\frac{(-1)^n}{n}+e^{\xi_n}\frac{1}{n^2}\right) \tag 1 \end{align}$$

we recall that the alternating harmonic series $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=\log 2$ converges. And we note that since $|e^{\xi_n}|<e$ for $0<|\xi_n|<\frac{1}{n}$, then the series

$$\begin{align} \sum_{n=1}^{\infty}e^{\xi_n}\frac{1}{n^2} &\le e\sum_{n=1}^{\infty}\frac{1}{n^2}\\\\ &=\frac{\pi^2\,e}{6} \end{align}$$

also converges. Finally, since the sum of two convergent series is a convergent series, the series on the left-hand side of $(1)$ converges.


PART 2: SHOWING CONVERGENCE IS CONDITIONAL

Now, we observe that the series of absolute values is bounded below as

$$\begin{align} \sum_{n=1}^{\infty}\left|e^{(-1)^n/n}-1\right|&=\sum_{n=1}^{\infty}\left|\frac{(-1)^n}{n}+e^{\xi_n}\frac{1}{n^2}\right|\\\\ &\ge \frac12\,\sum_{n=1}^{\infty}\frac{1}{n} \end{align}$$

which diverges since the harmonic series diverges. Thus, the series of interest is not absolutely convergent, only conditionally convergent.

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