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

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.

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  • $\begingroup$ @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
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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:

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

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

$$I_0(x)=\sum_{n=0}^\infty\frac{x^{2n}}{4^nn!^2}$$


Then we have that,

$$f(x,x)=\frac{e^{2x}-I_0(2x)}{2}$$

And in addition,

$$f(x,y)+f(y,x)=e^{x+y}-I_0(2\sqrt{xy})$$


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,

$$h(x,y)=\sum_{n=0}^\infty\frac{x^n}{n!}\sum_{k=0}^n\frac{y^k}{k!}=\sum_{n=0}^\infty\frac{x^n}{n!}\sum_{a+b=n}\frac{y^b}{b!}=\sum_{n=0}^\infty\sum_{a+b=n}\frac{x^{a+b}y^b}{(a+b)!b!}=\sum_{a=0}^\infty\sum_{b=0}^\infty\frac{x^{a+b}y^b}{(a+b)!b!}$$

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

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

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

$$h(x,y)=\sum_{n=0}^{\infty}\left(\frac{x^n}{n!}\sum_{r=0}^{n-1}\left(\frac{y^r}{r!}\right)\right)+\sum_{n=0}^\infty\frac{(xy)^n}{n!^2}=f(x,y)+I_(2\sqrt{xy})$$

Thus by simplification,

$$f(x,y)+f(y,x)=e^{x+y}-I_0(2\sqrt{xy})$$

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

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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 !

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    $\begingroup$ @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

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