Original posting by dioxen here: Double summation including power and factorial
I am finding some trouble in computing the following sum:
$$\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}$$
Could you please provide a result?
Thanks in advance
After a failed attempt at getting to the binomial expansion, I tried to play around with it a little bit. I came up with two ideas of a solution, but can't see it through. Both involve applying the Cauchy product.
\begin{equation}
\Bigg(\sum_{n=0}^\infty a_n \Bigg) \Bigg(\sum_{n=0}^{\infty} b_n \Bigg) = \sum_{n=0}^{\infty} c_n \space \text{ where } c_n = \sum_{k=0}^n a_k b_{n-k} \space \space \space \space \space \space \space \space \space \space \space \space (1)
\end{equation}
1: In this problem, we cound treat $$ c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!}=\frac{1}{n!}\sum_{k=0}^n\frac {(xy)^{n-k}}{(n-k)!}x^{k}=\frac{1}{n!}\sum_{k=0}^n a_k b_{ n-k}$$
So if we're thinking within the framework of the first two summations in (1), our $a_n$ and $b_n$ are as follows: $a_n=x^n$ and $ b_n = \frac {(xy)^{n}}{n!}\\$
our only problem is the we're left with the $\frac{1}{n!}$ in front of the summation. If we thought up some way to get rid of it, we'd have a nice geometric series and $e^{xy}$. But I can't think of a way…any leads here?
2: Alternatively, we could leave the $\frac{x^n}{n!}$ outside and treat it as a number (because it is, before we choose to sum it up over $n$). $$ c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!} 1^{n-k}=\frac{x^n}{n!}\sum_{k=0}^n a_k b_{ n- k}$$ which gives us $a_n=\frac{y^n}{n!}$ and $b_n=1$. Could we do something about $\frac{x^n}{n!}$?
$$
\Bigg(\sum_{n=0}^\infty \frac{x^n}{n!} \Bigg) \Bigg(\sum_{n=0}^{\infty} \frac{x^n}{n!} \frac{y^n}{n!} \Bigg)
$$
Does this even make sense? Because otherwise I can't think of a way this sum would be computable.
Best Answer
This function probably doesn't have a closed form, but here's how you can handle it. Define $$ f(x,y) = \sum_{0\leq m\leq k} \frac{x^k y^m}{k! m!}, $$ and simplify the inner sum using $$\sum_{0\leq m\leq k} \frac{y^m}{m!} = e^y \frac{k!-\gamma(1+k,y)}{k!} = e^y\left( 1 - \frac1{k!}\gamma(1+k,y)\right), $$ using equation (9) here: http://mathworld.wolfram.com/IncompleteGammaFunction.html. After that, we have $$ f(x,y) = e^{x+y} - e^y \sum_{k\geq0} \frac{x^k}{k!^2} \frac{y^{1+k}}{k+1} F\left(\begin{array}{c}k+1\\k+2\end{array}\middle| -y\right) = e^{x+y} - e^y h(x,y), $$ where $F$ is a hypergeometric function appearing in (7) on that page. Now the sum $h(x,y)$ can be simplified using the definition of a hypergeometric to get $$ h(x,y) = \sum_{k,l\geq0} \frac{x^k y^{1+k+l} (-1)^l}{k!^2 (k+1+l) l!}. $$ If we differentiate once w.r.t. $y$, the sum separates: $$ \partial_y h(x,y) = \sum_{k,l\geq0} \frac{x^k y^k}{k!^2} \sum_{l\geq0}\frac{(-y)^l}{l!} = I_0(2\sqrt{xy})e^{-y}, $$ where $I_0$ is a modified Bessel function.
Finally using $h(x,0)=0$, we can integrate $\partial_y h$ w.r.t. $y$ over $[0,y]$ to find that $$ f(x,y) = e^{x+y} - e^y \int_0^y e^{-y}I_0(2\sqrt{xy})\,dy. $$
It is very likely that that integral has no simple or closed form, as it doesn't appear in Gradshteyn and Ryzhik, but I could be wrong. This form is perhaps somewhat simpler than the above single sum with an incomplete gamma function.