1. ## Alternating Series

Determine the convergence or divergence of the series.
(E = sigma)

infinity
E (-1)^n+1 (2) / e^n - e^-n
n=1

2. i assume it is $\displaystyle \frac{2}{e^n - e^{-n}}$..

notice that it's a sequence of hyperbolic cosecant.. $\displaystyle \frac{2}{e^n - e^{-n}} = \frac{1}{sinh \,\, n} = csch \,\, n$

and from n=1 to infinity, that is monotone and bounded..

3. Originally Posted by element
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

4. yeah, i forgot to mention that $\displaystyle \lim_{n\rightarrow \infty} csch \,\, n = 0$
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