Trying to simplify the following sum: $$ \sum_{i=0}^n\frac{z^i}{(n-i)!}\,\frac{1}{(1+a)_i\,(1-a)_i}\sum_{j=0}^i(-1+a)_j\,(-1-a)_j\frac{(-z)^j}{j!}, $$ where $n=1,2,\ldots$, $z>0$, $0<a<1$, and $(x)_k=\Gamma(x+k)/\Gamma(x)$ which is the Pochhammer symbol.

The inner sum is expressible through the ${}_2F_{2}$ hypergeometric function. The following code

f[n_] := Sum[(z)^i/(n - i)!*1/(
Pochhammer[1 + a, i]*Pochhammer[1 - a, i])*
Sum[Pochhammer[-1 + a, j] Pochhammer[-1 - a, j]*(-z)^-j/j!, {j, 0,
   i}], {i, 0, n}];

f[k] // FullSimplify

gives the expression that involves the ${}_2F_{2}$ function.

My question is to what extent is the entire double sum above can be simplified? For example, can it be reduced to a single hypergeometric function of some sort?

  • $\begingroup$ Are you sure that this is a question about Mathematica the software? It seems to be entirely about the underlying math as you posed it. $\endgroup$ – MarcoB Feb 19 '19 at 3:33
  • $\begingroup$ As @MarcoB suggests, maybe try math.stackexchange.com $\endgroup$ – Roman Feb 19 '19 at 4:44
  • $\begingroup$ Do you have to have a simple algebraic expression for your sum? Because using your existing code, I get reasonable simple expressions when I plug integer values into your f[n] $\endgroup$ – Bill Watts Feb 20 '19 at 0:54
  • $\begingroup$ @BillWatts f[n] is a polynomial in z of degree n, so it’s not that bad. I was hoping to express f as a single hypergeometric function though. $\endgroup$ – Alex Feb 20 '19 at 1:44

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