Interval of convergence of $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$ I want to find the interval of convergence of $$\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$$
By the root test, $\sqrt[n]{|(\frac{n}{n+1})^{n^2}(2x)^n|}=|(\frac{n}{n+1})^{n}(2x)|$
And $(\frac{n}{n+1})^n=(1-\frac{1}{n+1})^n\to\frac1e$ as $n\to\infty$
So I've got an open interval(of convergence) s.t. $|\frac1e 2x|<1\implies |x|<\frac e2$.
Now I need to figure out whether the series is convergent or not at the end points, $x=\frac e2, -\frac e2$.
So there are two series I need to deal with, namely, $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}e^n$ and $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(-1)^ne^n$.
Suppose $(\frac{n}{n+1})^{n^2}$ is increasing. Then $$(\frac{n}{n+1})^{n^2}<(\frac{n+1}{n+2})^{(n+1)^2}\iff 1<(\frac{n+1}{n+2})^{(n+1)^2}(\frac{n+1}{n})^{n^2}$$
$$\iff 1<(1-\frac{1}{n+2})^{2n+1}(1+\frac{1}{n^2+2n})^{n^2}$$
But $(1-\frac{1}{n+2})^{2n+1}\to\frac{1}{e^2}$ and $(1+\frac{1}{n^2+2n})^{n^2}<(1+\frac{1}{n^2})^{n^2}\to e$
So I have $1<\frac1e<1$, which is a contradiction. Hence $(\frac{n}{n+1})^{n^2}$ is not increasing. But I don't know if it's decreasing. (is it true that $(1+\frac{1}{n^2+2n})^{n^2}\to e$?)
If $(1+\frac{1}{n^2+2n})^{n^2}\to e$ is true, then $(\frac{n}{n+1})^{n^2}e^n$ converges to some positive number, so $\sum(\frac{n}{n+1})^{n^2}e^n$ diverges. But I don't know if it's decreasing: can't apply the alternating series convergence test(Leibniz's criterion).
So, can you help me with the two end points?

One can use the Taylor expansion of $\log$ to get an equivalent of the sequence :
$$
\begin{align}
\left(\frac{n}{n+1}\right)^{n^2}e^n &= \exp{\left[n+n^2\log\frac{n}{n+1}\right]} \\
&= \exp{\left[n+n^2\left(-\frac{1}{n+1}-\frac{1}{2(n+1)^2}+o\left(\frac{1}{n^2}\right)\right)\right]} \\
&= \exp{\left[\frac{n^2+2n}{2(n+1)^2}+o(1)\right]} \\
&=\sqrt{e}+o(1).
\end{align}
$$
That should give you the information you need to conclude about the convergence of both your series.
