I am trying to find the sum $S$ of the following series. $$S = \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}t^{2n}\Gamma\left(\frac{1 + 2nH - H}{2H\left(1-H\right)}\right)}{\left(2n\right)!\rho^{\left(\frac{1 + 2nH - H}{2H\left(1-H\right)}\right)}} + i \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}t^{2n+1}\Gamma\left(\frac{1 + 2nH}{2H\left(1-H\right)}\right)}{\left(2n + 1\right)!\rho^{\left(\frac{1 + 2nH}{2H\left(1-H\right)}\right)}}$$
where $0 < H < 1$, $t\in \mathbb{R}$, and $\rho$ belongs to set of positive reals.
I have also been struggling to prove absolute convergence of this series as the Gamma function seems to get in the way with the ratio test, and I could not recall any similar series for comparison tests.
Any help or clues would be much appreciated. Thanks
PS : Perhaps ratio test along with the integral formula for the Gamma function and with L'Hopital's Rule might show the convergence. I'll try that out now.
The original question had $H\le 1$, but this causes a divide-by-zero problem, so I think only $H\in(0,1)$ is valid. Speaking of problems, the series doesn't converge. An key step is the following asymptotic of the Gamma function: if $g>0$, $$ \lim_{n\to\infty} \frac{\Gamma(gn+a)}{\Gamma(gn+b)}(gn)^{b-a}=1 $$In this problem we'll have $g=2/(1-H)$. I'll do the real part; the imaginary part works exactly the same. Using the Ratio Test and cleaning up, we have: $$ \lim_{n\to\infty}\left|\frac{t^{2 (n+1)} \rho ^{-\frac{2 H (n+1)-H+1}{(1-H) H}} \Gamma \left(\frac{2 (n+1) H-H+1}{(1-H) H}\right)}{(2 (n+1))!}\cdot \frac{(2 n)!\rho ^{\frac{2 H n-H+1}{(1-H) H}}}{t^{2 n} \Gamma \left(\frac{2 n H-H+1}{(1-H) H}\right)}\right| $$ $$ =t^2 \rho ^{-\frac{2}{1-H}}\lim_{n\to\infty}\left|\frac{ \Gamma \left(\frac{2 n}{1-H}+\frac{H+1}{H-H^2}\right)}{(2 n+1)(2n+2) \Gamma \left(\frac{2 n}{1-H}+\frac{1}{H}\right)}\right| $$Now use the asymptotic. $$ =t^2 \rho ^{-\frac{2}{1-H}}\lim_{n\to\infty}\left|\frac{1}{(2n+1)(2n+2)}\cdot\frac{ \Gamma \left(\frac{2 n}{1-H}+\frac{H+1}{H-H^2}\right)\cdot \color{blue}{\left(\frac{2}{1-H}n\right)^{\frac{2}{H-1}}}}{\Gamma \left(\frac{2 n}{1-H}+\frac{1}{H}\right)}\cdot \color{red}{\left(\frac{2}{1-H}n\right)^{\frac{2}{1-H}}}\right| $$ $$ =1\cdot t^2 \left(\frac{2}{\rho(1-H)}\right)^{\frac{2}{1-H}}\lim_{n\to\infty}\left|\frac{1}{(2n+1)(2n+2)}\cdot n^{2/(1-H)}\right| $$Since $0<H$, $2/(1-H)>2$, so this last term has a positive exponent in $n$. Thus the series diverges.