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

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.

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.

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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.