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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.

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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.

Comment:

  • 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.

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  • $\begingroup$ 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

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