Is it true, that $\sum\limits_{n=1}^{\infty}(e^{\frac{1}{n!e}}-1)=\frac{10}{11+e}$?

Question

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.

Answer 1

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.

Answer 2

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 $$0.632\approx \frac{1}{e}(e-1)<\sum\limits_{\color{red}{n=1}} \left(e^{\frac{1}{n!e}}-1\right)<\frac{1}{e}(e-1)e^{\frac{1}{e}}\approx 0.913$$