A closed form for $\sum\limits_n(e-(1+1/n)^n)$ I have been having some trouble trying to find a closed form for this sum. It seems to converge really slowly.

Find a closed form for $$S=\sum_{n=1}^\infty\left[e-\left(1+\dfrac{1}{n}\right)^n\right].$$

All I got so far is
$$
\begin{align}
e-\left(1+\dfrac{1}{n}\right)^{n} & = \sum_{k=0}^\infty\frac{1}{k!}
-\sum_{k=0}^n\binom{n}{k}\frac{1}{n^{k}} \\
& = \sum_{k=0}^\infty\frac{1}{k!}\left(1-\dfrac{n!}{(n-k)!}\dfrac{1}{n^k}\right) \\
& = \sum_{k=0}^\infty\frac{1}{k!}\left(1-\dfrac{(n)_k}{n^k}\right) \\
\end{align},
$$
$$
S=\sum_{k=0}^\infty\frac{1}{k!}\sum_{n=1}^\infty\left(1-\dfrac{(n)_k}{n^k}\right).
$$
Where $(n)_k$ is the Pochhammer symbol. But I don not know how I could carry on from here.
 A: Here's yet another approach, using the inequality $e^x\ge1+x$.
$e^x\ge1+x$ implies that $\log(1+x)\le x$. Thus,
$$
\begin{align}
\log\left(1+\frac1k\right)-\log\left(1+\frac1{k+1}\right)
&=\log\left(1+\frac1{k(k+2)}\right)\\
&\le\frac1{k(k+2)}\\
&=\frac12\left(\frac1k-\frac1{k+2}\right)
\end{align}
$$
Therefore,
$$
\begin{align}
n\log\left(1+\frac1n\right)
&=n\sum_{k=n}^\infty\left[\log\left(1+\frac1k\right)-\log\left(1+\frac1{k+1}\right)\right]\\
&\le\frac n2\sum_{k=n}^\infty\left(\frac1k-\frac1{k+2}\right)\\
&=\frac n2\left(\frac1n+\frac1{n+1}\right)\\
&=1-\frac1{2n+2}
\end{align}
$$
Exponentiating and applying $e^x\ge1+x$ yields
$$
\begin{align}
\left(1+\frac1n\right)^n
&\le e\cdot e^{-\frac1{2n+2}}\\
&\le\frac e{1+\frac1{2n+2}}\\
&=e\left(1-\frac1{2n+3}\right)
\end{align}
$$
Therefore,
$$
\bbox[5px,border:2px solid #00A0F0]{e-\left(1+\frac1n\right)^n\ge\frac e{2n+3}}
$$
Of course, this leads to the same conclusion: divergence of the series.
A: A more elementary answer to Jack's nice but perhaps complex answer involves looking at just the case $k=2$.
Writing $(n)_k=n(n-1)\cdots(n-(k-1))$, the falling factorial, we have:
$$\begin{align}
e-(1+1/n)^n &= \sum_{k=0}^\infty \frac{1}{k!}\left(1-\frac{(n)_k}{n^k}\right)\\
&\geq \frac{1}{2n}
\end{align}$$
since all the terms in the sum are positive, and $\frac{1}{2n}$ is the term when $k=2$.
A: $+\infty$ is a nice closed form.
By the Hermite-Hadamard inequality we have:
$$\log\left(1+\frac{1}{n}\right)^n = n\int_{n}^{n+1}\frac{dx}{x}\leq\frac{n}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)= 1-\frac{1}{2n+2}$$
hence, by the concavity of $1-e^{-x}$ over $\left[0,\frac{1}{4}\right]$:
$$ e-\left(1+\frac{1}{n}\right)^n \geq e\left(1-e^{-1/(2n+2)}\right)\geq\frac{4e}{2n+2}(1-e^{-1/4})\geq\frac{6}{5}\cdot\frac{1}{n+1}. $$
We can also prove that for any $n\geq 1$

$$ e-\left(1+\frac{1}{n}\right)^n \geq \frac{e}{2n+2}$$

holds. The conclusion is just the same.
A: Hint:
The first terms of the Taylor expansion of $e-(1+x)^{1/x}$ are
$$\frac{ex}2-\frac{e11x^2}{24}+\frac{e7x^3}{16}-\frac{e2447x^4}{5760}+\cdots$$
As there are no singularities around the origin, the series converges for $|x|\le\frac12$.
Summing with $x=\dfrac1n$, your series is asymptotic to the harmonic one.
