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.

  • $\begingroup$ I don't get it. Why don't you write $e^x$ instead of the power-series for it? $\endgroup$ – deufeufeu May 26 '13 at 9:48
  • $\begingroup$ @deufeufeu Because the two sums are dependent. The second one is FINITE! $\endgroup$ – shimee May 26 '13 at 12:48
  • $\begingroup$ ok I'm sorry. I misread the initial sum. $\endgroup$ – deufeufeu May 26 '13 at 13:07
  • $\begingroup$ I tried Laplace transform of it over $y$. Then I(Mathematica) could do the summation. I ended up with $\frac{s e^x-e^{x/s}}{(s-1) s}$ which I couldn't calculate the Inverse Laplace transformation of it. However, I can write that as an Integral, If that is good enough I will turn it to an answer. $\endgroup$ – Ali May 29 '13 at 18:45
  • $\begingroup$ @Ali sure, I'd be happy to see it (if you could also include some intermediate steps) $\endgroup$ – shimee May 29 '13 at 20:02

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.


We start by Laplace Transformation of the inner sum argument, this way we will get rid of the factorial in the denominator and end up with a simple series:

$$\mathcal{L}\left(\frac{y^m}{m!} \right)=s^{-(m+1)} $$

Where $\mathcal{L}$, stands for Laplace transform from $y$ to $s$(something tells me, this will deeply assume y is positive). Now we can do the inner summation easily:

$$\sum_{m=0}^k{s^{-(m+1)}}=\frac{s^{-k-1} \left(s^{k+1}-1\right)}{s-1}=\frac{1-s^{-k-1}}{s-1} $$

Now putting this back into the initial sum, we will have:

$$\sum_{k=0}^\infty \frac{x^k}{k!} \frac{1-s^{-k-1}}{s-1}=\frac{1}{s-1} \sum_{k=0}^\infty\left(\frac{x^k}{k!}-\frac{(\frac{x}{s})^k}{s\ k!} \right)$$

These sums are just the expansion of $e^z$, so we have:


Now, we have to calculate the inverse Laplace transform from $s$ to $y$. I could produce explicit expressions for all but the middle term(which I will write as an integral)

$$\sum_{k=0}^\infty \frac{x^k}{k!}\ \sum_{m=0}^k\frac {y^m}{m!}=\mathcal{L}^{-1}\left(\frac{e^x}{s-1}-\frac{e^{x/s}}{s-1}+\frac{e^{x/s}}{s} \right) \\ =e^{x+y}+J_0 (2\sqrt{xy})-\frac{1}{2 \pi i} \lim_{T\to\infty}\int_{ \gamma - i T}^{ \gamma + i T} e^{sy} \frac{e^{x/s}}{s-1}\,ds$$ where $\gamma$ is a real number so that the contour path of integration is in the region of convergence of $\frac{e^{x/s}}{s-1}$.

This is how far I could come. Hope it's a useful result for you.


Sadly, I believe there is no closed form for your double summation. The closest form I can get is $$\begin{eqnarray}\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!} &=& \sum_{k=0}^\infty \frac{x^k}{k!} \; {}_0F_1\left(\left.\begin{array}{c} - \\ k+1 \end{array} \; \right| xy \right) \\ &=& \frac 1{\pi}\sum_{k=0}^\infty \left(\frac xy \right)^{\frac k2} \; \int_0^{\pi} e^{2 \sqrt {xy} \cos(t)} \cos (kt) dt \end{eqnarray}$$

  • $\begingroup$ What is the expression $_0 F_1 ()$? Hypergeometric series? $\endgroup$ – shimee May 30 '13 at 13:41

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