Is it true, that $\sum\limits_{n=1}^{\infty}(e^{\frac{1}{n!e}}-1)=\frac{10}{11+e}$? If
$$\sum\limits_{n=1}^{\infty}(e^{\frac{1}{n!e}}-1)=\frac{10}{11+e}$$
is true, so how can we prove it (if not, how can we came to this approximation)?
If I made some mistakes, sorry for my English.
 A: Notice that if $h > 0$, then by the mean value theorem $e^h - 1 < e^h h$. So for any fixed integer $N \geq 1$ we have
\begin{align*}
\sum_{n=1}^{\infty} \left( e^{1/n!e} - 1 \right)
&< \sum_{n=1}^{N-1} \left( e^{1/n!e} - 1 \right) + e^{1/N!e} \sum_{n=N}^{\infty} \frac{1}{n!e} \\
&= \sum_{n=1}^{N-1} \left( e^{1/n!e} - 1 \right) + e^{1/N!e} \left(1 -
 \frac{1}{e} \sum_{n=0}^{N-1} \frac{1}{n!} \right).
\end{align*}
Now with the choice $N = 7$ with aid of numerical computation, we can check that
$$ \sum_{n=1}^{\infty} \left( e^{1/n!e} - 1 \right) < 0.7289542\color{red}{1703628121330\cdots} $$
Comparing this with
$$ \frac{10}{11+e} \approx 0.7289542\color{blue}{6155006217295\cdots} $$
it follows that the sum is strictly smaller than the proposed value.
A: This is not a complete answer, just an approximation.
From $\frac{x}{x+1}\leq \ln{(x+1)} \leq x, x>-1$ we can deduce that 
$\frac{x-1}{x}\leq \ln{x} \leq x-1, x>0$ or $\frac{e^x-1}{e^x}\leq x \leq e^x-1, \forall x$ and finally
$$x\leq e^x-1\leq xe^x$$
then
$$\frac{1}{n!e}\leq e^{\frac{1}{n!e}}-1\leq\frac{1}{n!e}e^{\frac{1}{n!e}}<\frac{1}{n!e}e^{\frac{1}{e}}$$
We know that $e=\sum\limits_{\color{red}{n=0}}\frac{1}{n!}$
taking partial sums and limits
$$0.632\approx \frac{1}{e}(e-1)<\sum\limits_{\color{red}{n=1}} \left(e^{\frac{1}{n!e}}-1\right)<\frac{1}{e}(e-1)e^{\frac{1}{e}}\approx 0.913$$
