# Trying to find the generic coefficient of a hypergeometric series

I begin with

FullSimplify[Series[HypergeometricPFQ[{-k,m/2-1,(m-1)/2,m/2},{1/2,1,3m/2-k-2},t],{t,0,3}],
Assumptions->k\[Element]Integers\[And]m\[Element]Integers\[And]k>=0\[And]m>0]


and obtain

Now I want to get a formula for the $$j$$th coefficient. Of course it is easy to guess it but then I should prove it, and also there are more complicated cases. So I try

FullSimplify[SeriesCoefficient[HypergeometricPFQ[{-k,m/2-1,(m-1)/2,m/2},{1/2,1,3m/2-k-2},t],{t,0,j}],
Assumptions->k\[Element]Integers\[And]m\[Element]Integers\[And]k>=0\[And]m>0]


The result (on 11.0.1.0) is zero. I would understand if the engine would reply that it cannot compute it, but why zero??

• It seems like the k>=0 assumption makes it zero. Are you sure that's correct? Aug 24, 2021 at 10:07
• @SjoerdSmit Thanks for this observation! I confirm that without the k>=0 I seem to obtain correct result. Which is even more strange since there are definitely nonzero cases with nonnegative k. You can observe this by computing the above series using, say, With[{k=3},...] Aug 24, 2021 at 11:42
• FullSimplify[ Piecewise[{{1/(-1 - k)!, j >= 0}}, 0], Assumptions -> k \[Element] Integers \[And] m \[Element] Integers \[And] k >= 0 \[And] m > 0] returns zero. So does Table[1/(-1 - k)!, {k, 100}]. Aug 24, 2021 at 12:15
• @MichaelE2 Well this is understandable. But in these coefficients, all negative factorials occasionally cancel out, how to deal with these cases? Aug 24, 2021 at 12:20
• Take the limits as k approaches an integer? However, Limit[(Gamma[j - k] Gamma[-2 - k + (3 m)/2])/(Gamma[-k] Gamma[-2 + j - k + (3 m)/2]), k -> k0, Assumptions -> k0 \[Element] Integers \[And] m \[Element] Integers \[And] m > 0 && j >= 0 && k0 >= 3] returns 0 even though the limits at specific values of j and m are generally nonzero. I'd be inclined to call the zero limit a bug, since it's not even generically true. Aug 24, 2021 at 14:41

HypergeometricPFQn[a_List, b_List, z_, n_] := z^n/n!