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