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.
Best Answer
$+\infty$ 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 $n\geq 1$
holds. The conclusion is just the same.