# Closed form for an infinite sum over Gamma functions?

I am having quite a bit of trouble trying to find a closed form (or a really fast way to compute) for the infinite sum $$\sum_{n=1}^{\infty} a^n \dfrac{\gamma(n+1,b)}{\Gamma(n+1)\Gamma(n)}$$ where $\gamma(n,b) = \int_0^bz^{n-1}e^{-z} dz$ is the lower incomplete Gamma function and $a,b > 0$.

Any ideas how to solve this?

Since: $$\sum_{n\geq 1}\frac{z^{n-1}}{\Gamma(n)\Gamma(n+1)}=\frac{I_1(2\sqrt{z})}{\sqrt{z}}$$ provided that $a,b\in\mathbb{R}^+$ you series equals: $$a \int_{0}^{b}\frac{I_1(2\sqrt{az})}{\sqrt{az}}e^{-z}\,dz = \int_{0}^{ab}\frac{I_1(2\sqrt{z})}{\sqrt{z}}e^{-z/a}=2\int_{0}^{\sqrt{ab}}I_1(2u)\,e^{-u^2/a}\,du.$$

• What is $I_1(z)$? – robjohn Apr 2 '15 at 11:19
• @robjohn: the modified Bessel function of the first kind. – Jack D'Aurizio Apr 2 '15 at 11:22
• @JackD'Aurizio: (+1) heh, I just figured that out while identifying the same function in this answer. – robjohn Apr 2 '15 at 12:07
• Why the downvote? – Jack D'Aurizio Apr 22 '15 at 20:22
• @Jack D'Aurizio Converting series back to the integral form is not the expected way of simplification. – Harry Peter Apr 24 '15 at 13:05

$\sum\limits_{n=1}^\infty\dfrac{a^n\gamma(n+1,b)}{\Gamma(n+1)\Gamma(n)}$

$=\sum\limits_{n=0}^\infty\dfrac{a^{n+1}\gamma(n+2,b)}{\Gamma(n+2)\Gamma(n+1)}$

$=\sum\limits_{n=0}^\infty\dfrac{a^{n+1}\int_0^bz^{n+1}e^{-z}~dz}{n!(n+1)!}$

$=\left[-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{a^{n+1}z^ke^{-z}}{n!k!}\right]_0^b$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\sum\limits_{n=0}^\infty\dfrac{a^{n+1}}{n!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{a^{n+1}b^ke^{-b}}{n!k!}$

$=ae^a-\sum\limits_{n=0}^\infty\dfrac{a^{n+1}b^{n+1}e^{-b}}{n!(n+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n+1}b^ke^{-b}}{n!k!}$

$=ae^a-abe^{-b}{_0F_1}(;2;ab)-\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{a^{n+1}b^ke^{-b}}{n!k!}$

$=ae^a-abe^{-b}{_0F_1}(;2;ab)-\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{a^{n+k+1}b^ke^{-b}}{(n+k)!k!}$

$=ae^a-abe^{-b}{_0F_1}(;2;ab)-ae^{-b}\Phi_3(1,1;a,ab)$ (according to http://en.wikipedia.org/wiki/Humbert_series)