<s>[INST] Series Convergence/Divergence $\frac{n^n}{(n+1)^{n+1}}$ Trying to establish whether $\sum x_n$ for $x_n := \frac{n^n}{(n+1)^{n+1}}$ converges or diverges. Here's what I've done so far:
1) n-th term: $x_n < \frac{n^n}{n^{n+1}} = \frac{1}{n}$, so $\lim(x_n) = 0$; n-th term test inconclusive. 
2) quotient test: $$
\begin{eqnarray*}
\lim \frac{x_{n+1}}{x_n} &=&\lim \frac{(n+1)^{n+1}/(n+2)^{n+2}}{n^n/(n+1)^{n+1}} \\
&=&\lim \frac{(n+1)^{2n+2}}{(n+2)^{n+2}n^n}\\
&=&\lim \left(\frac{n+1}{n+2}\right)^{n+2}\left(\frac{n+1}{n}\right)^{n}\\
&=& \lim \left(\frac{n+1}{n+2} \frac{n+1}{n}\right)^{n}\\
&=& \lim \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n}\\
&=& \lim \left(1+ \frac{1}{2n+n^2}\right)^{n}
\end{eqnarray*}
$$
I am not sure where to go from here. This looks a lot like the classic e limit $\lim (1+1/n)^n$, but it is sufficiently different that I cannot see how to transform it. 
3) Root Test:
$$
\begin{eqnarray*}
\lim x_n^{1/n} &=& \frac{n}{(n+1)^{1+1/n}} \\
&=& \lim \frac{n}{n+1} \frac{1}{(n+1)^{1/n}}
\end{eqnarray*}
$$
And again stuck here.
Thanks for your help,
Best wishes,
Leon
 [/INST] It's not hard to see that $(1+\frac{1}{n})^{-n}> \frac{1}{e}$, so $(1+\frac{1}{n})^{-n}\frac{1}{n+1}>\frac{1}{e}\frac{1}{n+1}$. But $\sum \frac{1}{e}\frac{1}{n+1}$ diverges so the original series diverges too.
</s>