Convergence of the series $\sum\limits_i\prod\limits_{j=1}^i\frac{j-2\alpha}{j+2\alpha}$ for different values of $\alpha$ I need help determining for which $\alpha$ values the following series converges (i.e. $\lim_{k\to\infty}a_k<\infty$):
$$a_k=1+\sum_{i=2}^k\prod_{j=1}^{i-1}\frac{j-2\alpha}{j+2\alpha}$$
I tried using convergence tests by haven't seen one that could help me. 
 A: By using the Pochhammer symbol $(x)_n = x\cdot(x+1)\cdots(x+n-1)$ we have
$$ a_k = 1+\sum_{i=2}^{k}\frac{(1-2\alpha)_{i-1}}{(1+2\alpha)_{i-1}}=1+\sum_{i=2}^{k}\frac{\Gamma(i-2\alpha)\Gamma(1+2\alpha)}{\Gamma(i+2\alpha)\Gamma(1-2\alpha)}=\sum_{i=1}^{k}\frac{\Gamma(i-2\alpha)\Gamma(1+2\alpha)}{\Gamma(i+2\alpha)\Gamma(1-2\alpha)}\tag{1} $$
hence:
$$\begin{eqnarray*} a_k &=& \frac{\Gamma(1+2\alpha)}{\Gamma(1-2\alpha)\Gamma(4\alpha)}\sum_{i=1}^{k}B(4\alpha,i-2\alpha)\\&=&\frac{\Gamma(1+2\alpha)}{\Gamma(1-2\alpha)\Gamma(4\alpha)}\sum_{i=1}^{k}\int_{0}^{1}x^{i-2\alpha-1}(1-x)^{4\alpha-1}\,dx\\&=&\frac{\Gamma(1+2\alpha)}{\Gamma(1-2\alpha)\Gamma(4\alpha)}\int_{0}^{1}(1-x^k)x^{-2\alpha}(1-x)^{4\alpha-2}\,dx\\&=&\frac{\Gamma(1+2\alpha)}{\Gamma(1-2\alpha)\Gamma(4\alpha)}\left[B(1-2\alpha,4\alpha-1)-B(k+1-2\alpha,4\alpha-1)\right]\\&=&\frac{2\alpha}{(4\alpha-1)}-\frac{\Gamma(1+2\alpha)\Gamma(k\color{red}{+1-2\alpha})}{\Gamma(k\color{red}{+2\alpha})\Gamma(1-2\alpha)(4\alpha-1)}\end{eqnarray*}$$
has a finite limit as $k\to +\infty$ iff $2\alpha\geq (1-2\alpha)$, i.e. iff $\color{red}{\alpha\geq\frac{1}{4}}$.
The limit in the case $\alpha=\frac{1}{4}$ is quite peculiar, given by $\frac{\gamma}{2}+\log(2)$.
