In my research I dealt to the following series. \begin{align*} \displaystyle\lim_{t\to\infty}\Big(\sum_{m=1}^{t}\Big(\frac{(\pi \xi)^{2m-1}}{(2m-1)!}\;\frac{1}{1-2^{\beta-2t+2m-2}}\Big)\Big) \end{align*} where $\beta\simeq0.7$ and $\xi\in\mathbb{C}$.
According some theorems I found that the limit exists. I tried to find the limit by comparing my series with the well-known series as follows: \begin{align*} \sum_{m=1}^{\infty}\frac{(\pi \xi)^{2m-1}}{(2m-1)!}=\sinh(\pi \xi). \end{align*} But the second fraction made it hard. Also, I found that the second fraction is bounded above. But my favourite is to find any well-known function or any closed form for my series. Can anyone help me. thanks.
$$ \begin{align} &\lim_{n\to\infty}\sum_{m=1}^n\frac{(\pi\xi)^{2m-1}}{(2m-1)!}\frac1{1-2^{\beta-2n+2m-2}}\\ &=\lim_{n\to\infty}\sum_{m=1}^\infty\frac{(\pi\xi)^{2m-1}}{(2m-1)!}\frac{[m\le n]}{1-2^{\beta-2n+2m-2}}\\ \end{align} $$ where $[\dots]$ are Iverson brackets.
Note that for $\beta\lt2$ $$ 0\le\frac{[m\le n]}{1-2^{\beta-2n+2m-2}}\le\frac1{1-2^{\beta-2}} $$ and $$ \lim_{n\to\infty}\frac{[m\le n]}{1-2^{\beta-2n+2m-2}}=1 $$ Thus, by Dominated Convergence, $$ \begin{align} &\lim_{n\to\infty}\sum_{m=1}^\infty\frac{(\pi\xi)^{2m-1}}{(2m-1)!}\frac{[m\le n]}{1-2^{\beta-2n+2m-2}}\\ &=\sum_{m=1}^\infty\frac{(\pi\xi)^{2m-1}}{(2m-1)!}\lim_{n\to\infty}\frac{[m\le n]}{1-2^{\beta-2n+2m-2}}\\ &=\sum_{m=1}^\infty\frac{(\pi\xi)^{2m-1}}{(2m-1)!}\\[3pt] &=\sinh(\pi\xi) \end{align} $$