$$ \sum_{n=0}^{\infty}\left(\frac{x^n}{n!}\sum_{r=0}^{n-1}\left(\frac{y^r}{r!}\right)\right) $$

This is a double summation I need to evaluate(not for a homework problem, but for a probability question involving gaming I found). I can't find any idea how to do this and wolfram alpha can't calculate this with extended computer time either, though this series is obviously less than e^(x+y) and monotonous, so it converges.

  • $\begingroup$ @Ethan: For $n=0$ the inner sum is $0$ (there are no $r$ with $0 \le r \le -1$). $\endgroup$ – Robert Israel Dec 29 '13 at 7:38
  • $\begingroup$ Are you interested in an explicit closed-form formula (which is unlikely to exist), or would an approximation suffice, possibly with an estimate of the error? In other words; are you more interested in calculating the value of the formula for given $x$ and $y$ with some precision, or do you need to manipulate it algebraically? $\endgroup$ – Peter Košinár Dec 29 '13 at 8:03

I will show you how to evaluate the special case for $x=y$ in terms of a modified Bessel function of the first kind. I will also give you a functional equation for your series in terms of the exponential function and a modified Bessel function of the first kind.

If we denote your function as:


And the modified Bessel function of the first kind with index zero as:


Then we have that,


And in addition,


To prove this we will define a slightly simpler function $h$ for convenience: $$h(x,y)=\sum_{n=0}^\infty\frac{x^n}{n!}\sum_{k=0}^n\frac{y^k}{k!}$$

Then we have that,


$$=\sum_{a=0}^\infty\sum_{b=0}^\infty\frac{y^bx^{a+b}}{b!(a+b)!}=\sum_{b=0}^\infty \frac{y^b}{b!}(\frac{x^{b}}{b!}+\frac{x^{1+b}}{(1+b)!}+\frac{x^{2+b}}{(2+b)!}+\frac{x^{3+b}}{(3+b)!}+\dots)$$

$$h(x,y)=\sum_{b=0}^\infty \frac{y^b}{b!}(\frac{x^{b}}{(b)!}+\frac{x^{1+b}}{(1+b)!}+\frac{x^{2+b}}{(2+b)!}+\frac{x^{3+b}}{(3+b)!}+\dots)$$ $$h(y,x)=\sum_{b=0}^\infty\frac{y^b}{b!}(\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+...\frac{x^{b-1}}{(b-1)!}+\frac{x^b}{b!}+\dots)$$

$$h(x,y)+h(y,x)=\sum_{b=0}^\infty \frac{y^b}{b!}(\frac{x^b}{b!}+e^x)$$ $$h(x,y)+h(y,x)=e^{x+y}+I_0(2\sqrt{xy})$$


Thus by simplification,


And the case for $x=y$ follows by simple algebraic manipulation.


The only thing I was able to find is the closed form expression of the inner summation. The result is given by $e^y \Gamma(n, y) / \Gamma(n)$ where $\Gamma(n, y)$ is the incomplete Gamma function. As you noticed, this is smaller than $e^y$ and the complete series is less than $e^{x+y}$.

I am not sure that a closed form could be found.

Happy New Year !

  • 1
    $\begingroup$ @Jason. Thanks for editing ! Happy New Year ! $\endgroup$ – Claude Leibovici Dec 29 '13 at 7:54
  • $\begingroup$ You're welcome ~ Happy New Year ~ $\endgroup$ – A. Chu Dec 29 '13 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.