<s>[INST] Prove or disprove convergence for the series: $\sum_{n=2}^{\infty}\left((1+\frac{1}{n})^n-e\right)^{\sqrt{\log(n)}}$ Because $(1+(1/n))^n$ converges to $e$, I was thinking of comparing the sum to the series $p^{\sqrt{\log(n)}}$, but this series apparently diverges according to wolfram so I'm at a loss. Can someone give me a clue as to where to go from here?
 [/INST] Note that
\begin{align}
\left(
1+\frac{1}{n}
\right)^{n}
=&
\frac{n!}{0!(n-0)!}\frac{1}{n^0}
+
\frac{n!}{1!(n-1)!}\frac{1}{n}
+
\cdots
+
\frac{n!}{n!(n-n)!}\frac{1}{n^n}
\\
=&
\frac{1}{0!}
+
\frac{1}{1!}\left( 1-\frac{1}{n}\right)
+
\frac{1}{2!}\left( 1-\frac{1}{n}\right)\left( 1-\frac{2}{n}\right)
+
\cdots
+
\frac{1}{n!}\left( 1-\frac{1}{n}\right)\cdot \cdots\cdot \left( 1-\frac{n-1}{n}\right)
\\
=&
\sum_{k=0}^{n}
\frac{1}{k!}
\prod_{i=0}^{k}
\left(
1-\frac{i}{n}
\right)
\end{align}
implies 
$$
\left| \;
 e-\left(1+\frac{1}{n} \right)^n 
\;\right|
=
\sum_{k>n}
\frac{1}{k!}
\prod_{i=1}^{k}
\left(
1-\frac{i}{n}
\right)
\leq
\sum_{k>n}
\frac{1}{k!}
=\frac{1}{n^2}
\sum_{k>n}
\frac{n^2}{k!}
$$
Here $\prod_{i=0}^{0}
\left(
1-\frac{k}{n}
\right)=1$. 
Since $n$ is constant with respect to $k$,
$
\sum_{k>n}
\frac{n^2}{k!}
$ is limited say by a constant $L$. Then we have
$$
\left| \;
 e-\left(1+\frac{1}{n} \right)^n 
\;\right|
<\frac{1}{n^2}\cdot L.
$$
Recall that $1 - \frac1x \leq \log x \leq x-1$ for all $ x > 0$. Then
\begin{align}
\sum_{n=2}^{\infty}
\left|e -
\left(
1+\frac{1}{n}
\right)^n
\right|^{\sqrt{\log(n)}}
&\leq
\sum_{n=2}^{\infty}
\left(\frac{L}{n^2}\right)^{\sqrt{\log(n)}}
\\
&\leq 
\sum_{n=2}^{\infty}
\left(\frac{L}{n^2}\right)^{\sqrt{1-1/n}}
\\
&\leq
\sum_{n=2}^{\infty}
\left(\frac{L}{n^2}\right)^{1/4}
\\
&=
\sum_{n=2}^{\infty}
\left(\frac{L}{n^{3/4}}\right)
\end{align}
</s>