Hypergeometric function evaluation 4F3 I need to show that for $m$ being non-negative integer,
the hypergeometric function ${}_4F_3$ below evaluates to $-1/2$ independent of $m$.
This is Mathematica notation, but we have 4 and 3 sets of parameters, and evaluate at $z=1$.
HypergeometricPFQ[{-1/3, 1/3, -3/2 - m, -m - 1}, {1/2, -2/3 - m, -1/3 - m}, 1]

I have looked in W. Koepf's book, but have not found any identity which can be specialized to this. Also, mathematica does not simplify this further.
As a side note, is there a nice list of hypergeometric identities in some good book?
I know Gasper and Rahman has a nice book, but this deals with the q-analogs mainly,
so I would prefer some reference which deals with the non-q-case.
 A: From https://dlmf.nist.gov/15.4 one has
$$
\sum_{k=0}^n\frac{(1/3)_k(-1/3)_k}{k!(1/2)_k}(-z^2)^k=\frac{1}{2}\left(\left(\sqrt{1+z^{2}}+%
z\right)^{2/3}+\left(\sqrt{1+z^{2}}-z\right)^{2/3}\right)
$$
and
$$
\sum_{k=0}^n\frac{(1/3)_k(2/3)_k}{k!(3/2)_k}(-z^2)^k=\frac{3}{2z}\left(\left(\sqrt{1+z^{2}}+%
z\right)^{1/3}-\left(\sqrt{1+z^{2}}-z\right)^{1/3}\right).
$$
Multiplying both sides and calculating coefficient of $(-z^2)^n$ one obtains, e.g. using $(x)_{n-k}=(-1)^k\frac{(x)_n}{(1-x-n)_k}$, that
$$
{}_4F_3\left({1/3,-1/3,-1/2-n,-n\atop 1/2,1/3-n,2/3-n};1\right)\cdot \frac{(1/3)_n(2/3)_n}{n!(3/2)_n}\\=-\frac{1}{2}\cdot\left[(-z^2)^n\right]\left\{\frac{3}{2z}\left(\left(\sqrt{1+z^{2}}+%
z\right)^{1/3}-\left(\sqrt{1+z^{2}}-z\right)^{1/3}\right)\right\},\quad n\in\mathbb{N},
$$
from which the claim follows (here $\left[(-z^2)^n\right]\left\{f(z)\right\}$ denotes the coefficient of $(-z^2)^n$ of the power series $f(z)$).
Gasper and Rahman is the best reference, even one is not working with $q$-series. Some other references, more suitable for combinatorists, might be Riordan's book, or Graham, Knuth, Patashnik's book (haven't read these two). Identities of this kind are called convolution type identites, because when written in terms of binomial coefficients the sum is a convolution.
