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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!

razivo
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  • Of course the term $i=0$ with $0$ denominator must be omitted. Hint: Evaluate $\sum_{i=1}^\infty \frac{x^i}{i!i}$ then substitute $x=1$. – GEdgar Sep 18 '21 at 20:12
  • We have $$ \sum_{n\ge 1}\frac{x^n}{n!n}=\sum_{n\ge 1}\int_0^1\frac{x^{n-1}}{n!} dx=\int_{0}^1\frac{e^x-1}{x},dx. $$ I don't know a closed form for the integral though. – Mastrem Sep 18 '21 at 20:16
  • Just an idea. (In your first attempt you try splitting in form $\frac{A}{i!}+\frac{B}{i}$). What if you tried (omitting first term of summation) $\frac{A}{(i-1)!}+\frac{B}{i^2}$ ? – kolobokish Sep 18 '21 at 20:16
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    @Mastrem You mean $\sum_n\int_0^x\frac{e^t-1}{t}dt$, which becomes your expression when $x=1$ as required. – J.G. Sep 18 '21 at 21:14
  • Note that it involves a limit of an exponential integral and natural logarithm. See the bolded link for the sum. We then have your integral $=\text{Ei}(1)-γ$ which is the Eulergamma/euler mascheroni constant. There are other special functions based on the link. – Тyma Gaidash Sep 19 '21 at 02:28

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