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I’ve come across this nested double sum while doing an investigation but cannot seem to find a closed form for it.

$$1+\sum_{i=1}^\infty{\frac{1}{i!} \sum_{j=0}^i}{\frac{1}{j!}}$$

This is about the simplest form I can get it to. I’m pretty sure it converges, using wolfram, to about 4.8343.

Some notes;
The closest I’ve managed to get is $e^{e-1}$, which is gotten when the top bound of the second sum is $\infty$ instead of $i$.
I think this may be some sort of convolution but I’m not sure which/how to undo that, I’ve studied generating functions a bit but not come across something like this.
For reference, it came up in a certain polynomial I’m looking at, namely $(1-x)^2(1-x/2)(1-x/6)(1-x/24) \cdots$
where the roots are factorial numbers.

Habeeb M
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1 Answers1

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Another way of looking at this expression:

Let your sum be $s=\sum_{i=0}^\infty\frac{1}{i!}\sum_{j=0}^i\frac{1}{j!}$. Then (summations are interchangeable because of absolute convergence):

\begin{align} s & = \sum_{i\geq0,\ 0\leq j\leq i}\frac{1}{i!}\frac{1}{j!}=\sum_{j\geq0,\ i\geq j}\frac{1}{i!}\frac{1}{j!}\\ & = \sum_{j\geq0}\frac{1}{j!}\sum_{i\geq j}\frac{1}{i!} \\ & = \sum_{j\geq0}\frac{1}{j!}(e-\sum_{0\leq i\leq j}\frac{1}{i!}+\frac{1}{j!}) \\ & = e^2-s+\sum_{j\geq0}\frac{1}{(j!)^2} \\ \implies s & = \frac{e^2+\sum_{j\geq0}\frac{1}{(j!)^2}}{2}. \end{align}

So your sum $s$ has a closed form (elementary) expression if and only if $\sum_{j\geq0}\frac{1}{(j!)^2}$ does. I am not personally sure if it does but I am guessing no, because Wolfram alpha tells me it is the modified Bessel function of first kind. Numerically, it equals approximately 2.2796. Therefore your sum evaluates to $$ s\approx\frac{e^2+2.2796}{2}\approx4.8343. $$ This matches your results.

trisct
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