Is it possible to express the following sum in terms of the hypergeometric function $_2F_1$:

$$ f(x) = \sum_{n=0}^\infty\frac{(-ax)^n}{n!~\Gamma(b-n)} $$

with $a$ and $b$ constant values ($x>0$ may also be treated as constant). If not, what other methods could be used?


For $|ax|<1$ the series converges to


We know it converges for $|ax|<1$ due to the ratio test:

$$ \lim_{n\to\infty}\Bigg|\left(\frac{(-ax)^{n+1}}{(n+1)!\Gamma(b-n-1)}\right)\left(\frac{n!\Gamma(b-n)}{(-ax)^n}\right)\Bigg|=|ax| $$ where I've used the fact that $\Gamma(b-n)/\Gamma(b-n-1)\sim -(n+1)$ as $n\to\infty$.

One way to evaluate the sum is to rearrange

\begin{align} \sum_{n=0}^\infty \frac{(-ax)^n}{\Gamma(b-n)n!}&=\frac{1}{\Gamma(b)}\sum_{n=0}^\infty \frac{\Gamma(b)}{\Gamma(b-n)}\frac{(-ax)^n}{n!}\\ &=\frac{1}{\Gamma(b)}\sum_{n=0}^\infty (b-1)_n\frac{(-ax)^n}{n!} \end{align} where $$ (x)_n = x(x-1)(x-2)...(x-n+1) $$ is the the falling factorial, and then use the exponential generating function $$ \sum_{n=0}^\infty (x)_n\frac{t^n}{n!}=(1+t)^x $$

  • $\begingroup$ Why the constraint $|ax| < 1$? $\endgroup$ – Brad Jul 15 '14 at 16:01
  • $\begingroup$ @ozo Could you please elaborate on how you calculated the sum? Clearly, it is not a geometric series as the common ratio depends on n. $\endgroup$ – aslan Jul 16 '14 at 9:41
  • $\begingroup$ @JDVlok I have updated my answer. $\endgroup$ – lemon Jul 16 '14 at 11:16
  • $\begingroup$ @ozo Thanks! It is much clearer now. $\endgroup$ – aslan Jul 16 '14 at 12:28

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