I was studying the entry Deceptively easy product from the Blog [1] and I then tried a variant:
Problem. Evaluate or calculate a get good approximation of $$\prod_{n=1}^\infty\left(1+\frac{\mu(n)}{a_n}\right),$$ where $$a_n=n!\sum_{k=1}^n\frac{\mu(k)}{k!}$$ and $\mu(n)$ is the Möbius function.
Then one can repeat the calculations (is an easy exercise from calculations of the author) to get $$\prod_{n=1}^N\left(1+\frac{\mu(n)}{a_n}\right)=\frac{2a_{N+1}}{(N+1)!},$$ since $a_2=2!\left(1-\frac{1}{2}\right)=1$.
Question. I would like to evaluate or give an approximation of our infnite product by means of this limit $$\lim_{N\to\infty}2\sum_{k=1}^N\frac{\mu(k)}{k!},$$ or with other method (see my attempt). Then I am asking what's about $$\sum_{k=1}^\infty\frac{\mu(k)}{k!}?$$ Many thanks all users.
My attempt. The absolute convergence of the series is obvious. Using an online calculator $\sum_{n=1}^{1000}\frac{\mu(k)}{k!}\approx 0.32619$. I've tried to use Abel's summation to get $$\lim_{N\to\infty}\sum_{k=1}^N\frac{\mu(k)}{k!}=\lim_{N\to\infty}\int_1^N\left(\sum_{1\leq k\leq t}\mu(k)\right)\frac{\Gamma'(t+1)}{\Gamma^2(t+1)}dt.$$ And my idea is try to combine with asymptotics, for Gamma or digamma function, inside the integral if it is feasible but I don't know how do it. Can you finish calculations to get an approximation of our product? By which way?
References:
[1] Ron's Blog RESIDUE THEOREM AND FRIENDS, www.residuetheorem.com.