Convergence of $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ Could somebody please check my solution?
I want to check, whether $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ converges or diverges.
Using the Comparison test:
Let $a_n = \frac{(1+\frac{1}{n})^n}{n^2},~ b_n=\frac{1}{n^2}$
Since $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$ converges and $\lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n}= \lim\limits_{n \rightarrow \infty} \frac{(1+\frac{1}{n})^n}{n^2} \frac{n^2}{1} = \lim\limits_{n \rightarrow \infty} (1+\frac{1}{n})^n = e$. 
Since $0<e<\infty$, $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ converges.
 A: Your solution is correct. However, the simpler trick implied by the problem is to notice the numerator. Recall that
$$
\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e
$$
Since all the terms are positive, we have 
$$
0\leq \sum_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^n}{n^2} \leq \sum_{n=1}^\infty \frac{e}{n^2}= e \sum_{n=1}^\infty \frac{1}{n^2}
$$
By the $p$-test, the series on the right converges. Therefore, your original series converges. 
EDIT As suggested by user21820, it might be unclear what I was saying. The limit just produces me a number, $e$ to use to bound the original series. The solution 'works' because $\left(1+\frac{1}{n}\right)^n<e$ for any positive integer $n$. This is usually discussed when one learns the definition of $e$. So then each term on the right sum is larger than that of corresponding ones in the original. The use of $e$ was arbitrary (but suggested by the numerator). We could have used 'any' number, $\pi, e^2,4,10,\sqrt{17}$, so long as $\left(1+\frac{1}{n}\right)^n$ was smaller than our choice of number, say $x$, for all positive integer $n$ (or at least all but finitely many of them). Then we would have written
$$
0\leq \sum_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^n}{n^2} \leq \sum_{n=1}^\infty \frac{x}{n^2}= x \sum_{n=1}^\infty \frac{1}{n^2}
$$
for whatever larger number we chose. 
A: Using the limit comparison test, 
$$ {1\over n^2}\left(1 + {1\over n}\right)^n \sim {e\over n^2},$$
your series converges.
