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?
 A: I will assume that the absolute value of $\left(1+\tfrac{1}{n}\right)^n-e$ is taken since otherwise the terms don't exist, as mentioned in the comment.
The difference between $e$ and $\left(1+\tfrac{1}{n}\right)^n$ is of order $\tfrac{1}{n}$ (expand the binomial to the second order term to see that; this is mentioned in the comments, too). So for large $n$ the terms are of order $n^{-\sqrt{\log(n)}}$. For $n$ large enough, they are bound from above by, say, $n^{-2}$; hence, the series converges.
A: Hint: Prove that
$$\lim_{n\to\infty} \frac{e-\left(1+\frac{1}{n}\right)^n}{\frac{e}{2n}} = 1.$$
A: 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}
A: Do you mean $\enspace\displaystyle\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^{\sqrt{\ln n}}\enspace$ ? 
If Yes : See $\enspace\displaystyle\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2\enspace$ (convergent) in 
Closed form for $\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2$? .
We have $\enspace\displaystyle 0<\sum_{n=55}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^{\sqrt{\ln n}}<\sum_{n=55}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2\enspace$ and $\left(\left(e-\left(1+\frac{1}{n}\right)^n \right)^{\sqrt{\ln n}}\right)_n$ is a null sequence therefore we have convergence.  
A: In this answer it is shown that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing. Therefore,
$$
\left(1+\frac1n\right)^n\le e\le\left(1+\frac1{n-1}\right)^n\tag{1}
$$
Using the formula for $\frac{a^n-b^n}{a-b}$, we get
$$
\begin{align}
\frac{\left(1+\frac1{n-1}\right)^n-\left(1+\frac1n\right)^n}{\frac1{n-1}-\frac1n}
&=\sum_{k=1}^n\left(1+\frac1{n-1}\right)^{n-k}\left(1+\frac1n\right)^{k-1}\\[6pt]
&\le n\left(1+\frac1{n-1}\right)^{n-1}\\[12pt]
&\le ne\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
e-\left(1+\frac1n\right)^n
&\le\left(1+\frac1{n-1}\right)^n-\left(1+\frac1n\right)^n\\
&\le\frac{e}{n-1}\tag{3}
\end{align}
$$

$$
\begin{align}
\sum_{n=2}^\infty\left(e-\left(1+\frac1n\right)^n\right)^{\sqrt{\log(n)}}
&\le\left(e-\frac94\right)^{\sqrt{\log(2)}}+\sum_{n=3}^\infty\left(\frac{e}{n-1}\right)^{\sqrt{\log(n)}}\\
&\le\left(e-\frac94\right)^{\sqrt{\log(2)}}+\sum_{n=3}^\infty\left(\frac{e}{n-1}\right)^{\sqrt{\log(3)}}\tag{4}
\end{align}
$$
where the sum converges because $\sqrt{\log(3)}\gt1$.
