1
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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

$$\frac{(1-ax)^{b-1}}{\Gamma(b)}$$


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 $$ \sum_{n=0}^\infty (x)_n\frac{t^n}{n!}=(1+t)^x $$

$\endgroup$
  • $\begingroup$ 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

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.