Evaluate $\sum_{i=1}^\infty \frac{1}{i!i}$
Attempt 1 - partial fractions:
$$\frac{1}{i!i}=\frac{A}{i!}+\frac{B}{i}$$
$$1=iA+i!B$$
which is impossible without creating a denominator in $A$ which defeats the purpose.
Attempt 2 - Gamma function:
$$\sum_{i=1}^\infty \frac{1}{i!i}=\sum_{i=1}^\infty \frac{1}{\Gamma(i+1)i}=\sum_{i=1}^\infty \frac{\Gamma(i)}{\Gamma(i+1)^2}$$
Which isn't very promising since it isn't a beta function.
I thought about using leibnez's method or turning it into an integral but haven't managed.
If you can help, I'd like if you provided a hint and the solution in a spoiler box.
thanks!