Convergence of weighted series of Gamma functions I am interested in determining if the following series converges:
$$\sum_{k=0}^{\infty} \frac{2^{2\delta}}{\Gamma(2\delta)} \frac{\Gamma(k+2\delta)}{\Gamma(k+1)} \frac{1}{(k+\delta)^2} \;,$$
where $\delta > 0$ is a real number and the gamma function is defined by $$\Gamma(\alpha) = \int_0^{\infty} x^{\alpha-1} e^{-x} \, dx \;,$$ for $\alpha > 0$. Any thoughts on this would be much appreciated.
Related question / more thoughts:
Define $-2\delta \choose k$ $= \frac{(-1)^k \Gamma(k+2\delta)}{\Gamma(k+1)\Gamma(2\delta)}.$ Does the following series converge?
$$\sum_{k=0}^\infty 2^{2\delta} {-2\delta \choose k} \frac{1}{(k+\delta)^2} $$
I've been trying to reconcile these two series with Newton's generalized binomial theorem, which states that $$\sum_{k=0}^\infty {-2\delta \choose k} x^k = (1+x)^{-2\delta}$$ for $|x|<1$ and $\delta \in \mathbb{R}$.
 A: Use Stirling's approximation to get the following asymptotic behavior:
$$ \sqrt{\frac{k+2\delta-1}{k}} \left(\frac{k+2\delta-1}{k}\right)^k \left(\frac{k+2\delta-1}{e}\right)^{2\delta-1}\frac{1}{(k+\delta)^2}$$
Automatically from this expression we have that the series diverges if $\delta \geq \frac{3}{2}$ so all that remains to be seen is the behavior when $\delta\in \left(0,\frac{3}{2}\right)$. Limit compare this expression with a series with known convergence or divergence, such as $\frac{1}{(k+\delta)^{3-2\delta}}$:
$$\lim_{k\to\infty} \sqrt{\frac{k+2\delta-1}{k}} \left(1+\frac{2\delta-1}{k}\right)^k \left(\frac{k+2\delta-1}{e}\right)^{2\delta-1}\frac{1}{(k+\delta)^{2\delta-1}}$$
$$ = 1\cdot e^{2\delta-1}\cdot e^{1-2\delta}\cdot\lim_{k\to\infty}\left(1+\frac{\delta-1}{k+\delta}\right)^{2\delta-1} = 1$$
If $\delta < 1$, $\frac{1}{(k+\delta)^{3-2\delta}}$ converges, but for $\delta \geq 1$ it diverges. Thus the series shares the same behavior and converges when $0 <\delta < 1$.
A: The ratio test is also interesting to analyze. Let
$$|a_k|=\frac{2^{2 \delta }\,\Gamma (k+2 \delta )}{\Gamma (2 \delta )
 \,  \Gamma (k+1) \,(\delta +k)^2}$$ Take logarithms and use Stirling approximation. Apply it twice and continue with Taylor series
$$\left|\frac{a_{k+1}}{a_k}\right|=e^{\log(|a_{k+1}|)-\log(|a_{k}|)}=1+\frac{2 \delta -3}{k}+O\left(\frac{1}{k^2}\right)$$
Around $\delta=0$, the summation is 
$ \sim  \frac{1}{\delta^2}$ and does not converge if $\delta=0$.
For $\delta=1$, we have $a_k=4\frac{ (-1)^k}{k+1}$ and the summation  converges to $4 \log (2)$
For $\delta=\frac 32$, we have $a_k=\frac{(-1)^k16  (k+1) (k+2)}{(2 k+3)^2}$ and the summation does not converge.
