I'm trying to figure out the conditions for $$\sum_k^{\infty} (N+k)a^{2k}{}_2F_1\left(1-k,1-N-k;2;\left(\frac{b}{a}\right)^2\right)$$ converging, where $N\in\mathbb{N}$, $\{a,b\}\in \mathbb{R}^+$ and ${}_2F_1(a,b;c;z)$ is the Gaussian hypergeometric function.
Thanks to @gammatester, the Hypergeometric function reduces to a polynomial, so the sum can be written as
$$\sum_{k=1}^{\infty} (N+k)a^{2k} \sum_{n=0}^{k-1}(-1)^n\binom{k-1}{n}\frac{(1-N-k)_n}{(2)_n}\left(\frac{b}{a} \right)^{2n},$$
where the Pochhammer symbol $(x)_n$ has been used.
Looking for a moment only at the exponential factors and considering only the biggest value of $n$ $(n=k-1)$, we have something that scales as $$a^{2k}\left(\frac{b}{a}\right)^{2(k-1)}=a^2b^{2k-2},$$
so I think $b<1$ is required. But from doing numerical analysis in Mathematica, I found $a<1$ and $a+b\leq 1$ also seems to be necessities.
Is this true? If yes, why? And are there any other constraints for the sum to converge?
Thanks.