I have the following sum \begin{equation} \sum_{n=1}^{a-1} \frac{(1-a)_{n}}{(2-b)_{n}}\,\frac{\Gamma(n)}{n!}\left(\frac{d}{c}\right)^{n}, \end{equation} where $a=1,2,3,\dots$, $b=2,3,4,\dots$, $c>0$, $d>0$, and $(s)_{n}$ is the Pochhammer symbol. Looking at this sum I can see that it is very similar to the definition of $_{1}F_{1}(1-a;2-b;\tfrac{d}{c})$ where \begin{equation} {_{p}F_{q}}(a_{1},a_{2},\dots,a_{p};\,b_{1},b_{2},\dots,b_{q};\,z) = \sum_{n=0}^{\infty}\frac{(a_{1})_{n}(a_{2})_{n}\cdots (a_{p})_{n}}{(b_{1})_{n}(b_{2})_{n}\cdots (b_{q})_{n}}\ \frac{z^{n}}{n!}, \end{equation} is the generalized hypergeometric function. Unable to put the sum in terms of a hypergeometric function, I tried typing it into WolframAlpha here and got the following answer: \begin{equation} \frac{(a-1)d}{(b-2)c}\,{_{3}F_{2}}\left(1,1,2-a;\,2,3-b;\,\tfrac{d}{c}\right). \end{equation}

Does anyone see how this answer can be derived?

I can see how pieces of the answer make sense, e.g. that the solution does truncate to $a-1$ terms because of the $2-a$ in the first argument of the $_{3}F_{2}$ function, but I am unable to fully reproduce the answer.


We obtain \begin{align*} \color{blue}{\frac{(a-1)d}{(b-2)c}}&\color{blue}{\,{_{3}F_{2}}\left(1,1,2-a;\,2,3-b;\,\tfrac{d}{c}\right)}\\ &=\frac{(a-1)d}{(b-2)c}\sum_{n=0}^\infty\frac{(1)_n(1)_n(2-a)_n}{(2)_n(3-b)_n}\,\frac{1}{n!}\left(\frac{d}{c}\right)^n\tag{1}\\ &=\sum_{n=0}^\infty\frac{n!n!(1-a)_{n+1}}{(n+1)!(2-b)_{n+1}}\,\frac{1}{n!}\left(\frac{d}{c}\right)^{n+1}\tag{2}\\ &=\sum_{n=0}^\infty\frac{(1-a)_{n+1}}{(2-b)_{n+1}}\,\frac{\Gamma(n+1)}{(n+1)!}\left(\frac{d}{c}\right)^{n+1}\tag{3}\\ &=\sum_{n=1}^\infty\frac{(1-a)_{n}}{(2-b)_{n}}\,\frac{\Gamma(n)}{n!}\left(\frac{d}{c}\right)^{n}\tag{4}\\ &\color{blue}{=\sum_{n=1}^{a-1}\frac{(1-a)_{n}}{(2-b)_{n}}\,\frac{\Gamma(n)}{n!}\left(\frac{d}{c}\right)^{n}}\tag{5}\\ \end{align*} and the claim follows.


  • In (1) we use the definition of hypergeometric functions.

  • In (2) we multiply out and use factorials instead of the Pochhammer symbol.

  • In (3) we cancel terms and simplify.

  • In (4) we shift the index to start with $n=1$.

  • In (5) we observe that terms with index $\geq a$ are zero.

  • $\begingroup$ Thank you for the complete and well written answer. $\endgroup$ – Aaron Hendrickson Sep 21 '17 at 10:21
  • $\begingroup$ @AaronHendrickson: You're welcome! :-) $\endgroup$ – Markus Scheuer Sep 21 '17 at 11:05

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.