I have a few problems concerning infinite series which I need some help on. I'm supposed to determine the convergence/divergence using ONLY the comparison tests. Here are the problems:

$\displaystyle \displaystyle\Sigma\frac{1}{e^n+1}$

$\displaystyle \displaystyle\Sigma\frac{2^n}{3^n-1000}$

$\displaystyle \displaystyle\Sigma\frac{1}{e^n^2}$

For the first and the third ones, I want to apply the plain comparison test, using $\displaystyle \frac{1}{x^2}$ as the series which I compare it to. However, I need a way to prove that $\displaystyle e^n$ is greater than $\displaystyle x^n$ for all n [1,$\displaystyle \infty)$. Once I prove this, how would I apply it to the third problem since it's $\displaystyle {e^{n}}^2$? By the way, that is e^n^2...it's kind of hard to tell with the latex. For the second problem, here is what I have so far:

$\displaystyle \displaystyle\Sigma\frac{2^n}{3^n-1000}$

$\displaystyle \displaystyle\frac{2^n}{3^n-1000}\geq\frac{2^n}{3^n}$

So, now using the limit comparison test...

$\displaystyle \displaystyle\lim_{n\to\infty}\frac{\frac{2^n}{3^n-1000}}{\frac{2^n}{3^n}} $ = $\displaystyle \displaystyle\lim_{n\to\infty}\frac{3^n}{3^n-1000}$=1

The limit exits, is finite, and is greater than zero...so the original series will behave the same as $\displaystyle \Sigma\frac{2^n}{3^n}$ but I'm not sure how to prove whether this series converges/diverges? Any help would be appreciated.