Let $\mu(n)$ the Möbius function, and for integers $k\geq 0$ let $B_k$ the Bernoulli numbers, see the definition from this MathWorld.
Combining the Prime Number Theorem with the generating function for Bernoulli numbers I wrote $$0=\sum_{k=1}^\infty \mu(k)e^{1/k}\sum_{n=1}^\infty\frac{B_n}{n!}\frac{1}{k^n}-\sum_{n=2}^\infty\frac{B_n}{n!}\frac{1}{\zeta(n)}+\sum_{k=1}^\infty\mu(k)\left(e^{1/k}-1\right).\tag{1}$$
I know that how to discuss that the last two series of $(1)$ are convergent, the series in the middle has a closed-form ($B_n$=0 for odd integers $n>1$ and there are a well-known relation between the particular values at even integers $\geq 2$ of $\zeta(n)$ and $B_n$) and using for the third series I know how to discuss the convergence using the Taylor series around the origin $x=0$ of $e^(x)-1$, and again the Prime Number Theorem and particular values of some Dirichlet series.
As a consequence the first series in $(1)$ is also convergent. I would like to know a direc proof of this fact.
Question. How do you prove that (I write it as is showed below) $$\sum_{k=1}^\infty \sum_{n=1}^\infty\frac{B_n}{n!}\frac{\mu(k)e^{1/k}}{k^n}$$ does converge? Many thanks.
Using $$\sum_{n=0}^\infty B_n\frac{x^n}{n!}=\frac{x}{e^x-1}$$ for $|x|<2\pi$, the inner sum is $$\frac{1+1/k-e^{1/k}}{e^{1/k}-1}$$ and this is $$-\frac1{2k}+O(k^{-2}).$$ As $e^{1/k}=1+O(1/k)$ as $k\to\infty$, your overall sum is $$\sum_{k=1}^\infty\mu(k)\left(-\frac{1}{2k}+O(k^{-2})\right) =-\frac12\sum_{k=1}^\infty\frac{\mu(k)}k+\sum_{k=1}^\infty O(k^{-2}).$$ The last sum converges, and it's a consequence of the Prime Number Theorem that $\sum_{k=1}^\infty\mu(k)/k$ converges (to zero).