Determine the convergence or divergence of the series.
(E = sigma)
infinity
E (-1)^n+1 (2) / e^n - e^-n
n=1
Lets put this in math notation:
$\displaystyle
\sum_{n=1}^{\infty} (-1)^{n+1} \frac{2}{e^n-e^{-n}}
$
Well for large $\displaystyle n$, $\displaystyle \frac{2}{e^n-e^{-n}}$ is strictly decreasing and goes to zero, and so the alternating series test applies and your series is convergent.
RonL