Information on the sum $\sum_{n=1}^\infty \frac{\log n}{n!}$
In my personal study of interesting sums, I came up with the following sum that I could not evaluate:
$$\sum_{n=1}^\infty \frac{\log n}{n!} = 0.60378\dots$$
I would be very interested to see what can be done to this sum. Does a closed form of this fascinating sum exist?
Solution 1:
Using Dobinski's formula for Bell numbers, we have $$B(n)=\frac{1}{e}\sum_{k=0}^{\infty}\frac{k^n}{k!}$$ Hence, $$\frac{d}{dn}B(n)=\frac{1}{e}\sum_{k=2}^{\infty}\frac{k^n\log k}{k!}$$ whence, $$\sum_{k=1}^{\infty}\frac{\log k}{k!}=B'_0 e$$ Note that the first term ($k=1$) is $0$.