I have seen hypergeometric functions over the years on Wolfram Alpha and am trying to learn more about them. I recently read this question and its associated answers, but understood very little. I wrote a program to arbitrarily search for interesting hypergeometric function values and stumbled on this one.
How do I prove the following identity?
$$ {_2F_1}\left(-\frac{19}{20}, \frac{11}{30}; -\frac{19}{30}; -2\right) = 0 $$
This would be equivalent to the following sum, where $(x)_n$ refer to the rising factorial or Pochhammer symbol. However, $\left|- 2\right|$ is not less than $1$, so this sum is not guaranteed to converge.
$$ \sum_{n=0}^{\infty} \frac{\left(-\frac{19}{20}\right)_n \cdot \left(\frac{11}{30}\right)_n}{\left(-\frac{19}{30}\right)_n\cdot(1)_n} \cdot (-2)^n $$
I tried applying the first Pfaff transformation in order to get $-2$ back in the radius of convergence:
$$ {_2 F_1}(a, b; c; x) \Longrightarrow (1-z)^{-b}\cdot{_2 F_1}\left(b,c-a;c;\frac{z}{z-1}\right) $$
$$ {_2F_1}\left(-\frac{19}{20}, \frac{11}{30}; -\frac{19}{30}; -2\right) \Longrightarrow 3^{-\frac{11}{30}} \cdot {_2F_1} \left( \frac{11}{30}, \frac{19}{60}; -\frac{19}{30}; \frac{2}{3} \right) $$
Because the sum converges to zero, I can ignore the leading $3^{-\frac{11}{30}}$ .
The trick worked and it gave me something I can sum numerically.
Here is a table with the first 10 terms of the transformed hypergeometric series (without the leading constant).
0 1.0
1 -0.12222222222222222
2 -0.19993827160493827
3 -0.1782466849565615
4 -0.14016354150790022
5 -0.1046338569817722
6 -0.07596678344256204
7 -0.05421630175119416
8 -0.03824906471494405
9 -0.026761952441104003
And here are the first twenty partial sums
0 0.0
1 1.0
2 0.8777777777777778
3 0.6778395061728395
4 0.499592821216278
5 0.3594292797083778
6 0.2547954227266056
7 0.17882863928404355
8 0.1246123375328494
9 0.08636327281790535
10 0.05960132037680134
11 0.040992463681377815
12 0.028115033171369225
13 0.01923797177189061
14 0.01313772602200051
15 0.008956592581665427
16 0.006097117468555947
17 0.004145193056747411
18 0.00281493829147749
19 0.0019096402360084949
At this point, however, I'm stuck. The values in the transformed series are not particularly friendly and I don't see an obvious way to bound the partial sums.
Best Answer
Your function is of the form $ _2 F_1(a,b;b-1;z) $. We can cancel $(b)_n / (b-1)_n = \frac{n}{b-1}-1$, and then note that $ _2 F_1 (a,b;b;z) = (1+z)^a $. After cancellation, the general case is $$ _2 F_1(a,b;b-1;z) =\frac{ (a-b+1)z+(b-1)}{(b - 1) (1-z)^{a+1}} $$In particular, your function is $\displaystyle{\frac{z+2}{2(1-z)^{1/20}}}$, which evaluates to zero at $z=-2$.