what is $\sum_{k=1}^{\infty} e^{\frac{1}{k}}-1-\frac{1}{k}$?

Question

I tried to use $e^{x}=\sum_{i=1}^{\infty} \frac{x^{i}}{i!}$ and then change the order of summation but it didn't work out. I also didn't manage to prove that it converges, so I would be really grateful if someone could sort that out for me

Answer 1

First of all, for all $x$, $e^x \geqslant x + 1$ (to prove it, study the variations of the function $x \mapsto e^x - x - 1$) so the terms are non negative and $e^x - x - 1 = \sum_{n \geqslant 2} \frac{x^n}{n!} \sim \frac{x^2}{2}$ when $x \rightarrow 0$ thus $e^{1/k} - \frac{1}{k} - 1 \sim \frac{1}{2k^2}$, which implies that your series converges by Riemann criterion. Let us call $S$ the value of the sum.

Then, you have $e^{1/k} - \frac{1}{k} - 1 = \sum_{n \geqslant 2} \frac{1}{k^nn!}$ and since all the terms are non negative, you can use Fubini theorem to get that, $$ S = \sum_{k \geqslant 1} e^{1/k} - \frac{1}{k} - 1 = \sum_{n \geqslant 2}\sum_{k \geqslant 1} \frac{1}{k^nn!}. $$ Now, for all $n \geqslant 2$, $\sum_{k \geqslant 1} \frac{1}{k^n} = \zeta(n)$ so $$ S = \sum_{n \geqslant 2} \frac{\zeta(n)}{n!}. $$ And Wolfram alpha gives $S \approx 1.07569$ but I doubt there exists a close form. If you want to express $S$ as an integral for example, you can use the Abel's summation formula (https://en.wikipedia.org/wiki/Abel%27s_summation_formula) that implies that for all $s$ such that $\Re(s) > 1$, $\zeta(s) = s\int_1^\infty \frac{\lfloor t\rfloor}{t^{s + 1}} \, dt$. Once more, all the integrands/terms are non negtaive so we can use Fubini theorem to get, \begin{align*} S & = \int_1^\infty \sum_{n \geqslant 2} \frac{n\lfloor t\rfloor}{n!t^{n + 1}} \, dt\\ & = \int_1^\infty \frac{\lfloor t\rfloor}{t^2}\sum_{n \geqslant 2} \frac{1}{(n - 1)!t^{n - 1}} \, dt\\ & = \int_1^\infty \frac{\lfloor t\rfloor}{t^2}\left(e^{1/t} - 1\right) \, dt\\ & = \int_0^1 \left\lfloor\frac{1}{x}\right\rfloor(e^x - 1) \, dx \textrm{ with $x = 1/t$.} \end{align*} Once more, I doubt there is a closed form.