Convergent series and comparison test Show that the series converges, but not absolutely: $\displaystyle \sum_{n=1}^\infty\Bigg(\exp\Bigg(\frac{(-1)^n}{n}\Bigg)-1\Bigg)$.
This is what I did so far:
$\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1=\exp(\frac{1}{n})-1>0$ when $n$ is even
$\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1=\exp(\frac{-1}{n})-1<0$ when $n$ is odd
So, $\Bigg|\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1\Bigg|\leq\exp(\frac{1}{n})$. I was going to use comparison test but now stuck. Any hints?

Use the fact that $\exp(x) = 1 + x + x^2/2! + x^3/3!+ \cdots$ for any $x$, so
$$
\exp(\frac{(-1)^n}{n}) = 1 + \frac{(-1)^n}{n} + \frac{1}{2!}\left(\frac{(-1)^n}{n}\right)^2 + \dots.
$$
Then 
$$
\Bigg|\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1\Bigg|\leq\frac{2}{n} \to 0.
$$
Indeed
$$
\Bigg|\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1\Bigg|\leq\frac{1}{n} + \Bigg|\frac{1}{2!}\left(\frac{(-1)^n}{n}\right)^2 + \frac{1}{3!}\left(\frac{(-1)^n}{n}\right)^3 + \dots\Bigg| 
\leq \frac{1}{n} + \Bigg|\left(\frac{1}{n}\right)^2 + \left(\frac{1}{n}\right)^3 + \dots\Bigg| \leq \frac{1}{n} + \frac{1}{n^2}\frac{n}{n-1} \leq \frac{2}{n}.
$$
Let's prove monotony. I'm going to show that $|a_{2n+1}|<|a_{2n}|<|a_{2n-1}|$.
For $2n$ we have
$$
|a_{2n}| = \frac{1}{2n} + \frac{1}{2!}\left(\frac{1}{2n}\right)^2 +  \frac{1}{3!}\left(\frac{1}{2n}\right)^3 + \dots
$$
Then
$$
|a_{2n+1}| \leq\frac{1}{2n+1} + \frac{1}{2!}\left(\frac{1}{2n+1}\right)^2 +\dots < \frac{1}{2n} + \frac{1}{2!}\left(\frac{1}{2n}\right)^2 +\dots < \frac{1}{2n-1}+\frac{1}{2}\frac{1}{(2n-1)^2} \leq |a_{2n-1}|.
$$
Now by Leibniz's test the series converges.
As mvggz noted the series doesn't converge absolutely because $|a_n|\sim \frac{1}{n}$.
