Let $(S_n, n\geq0$) be a simple random walk with $S_0=0$ and define $T:=inf\{n\geq 1 : S_n=0\}$. In class we proved that $$\mathbb{P}(T=2n)=\dfrac{1}{2n-1}\binom{2n}{n}2^{-2n}.$$ The professor then said that using this and Stirling Approximation($n!\sim n^n(\sqrt{2\pi n})e^{-n}$) one can prove that $$\mathbb{E}(T^\alpha)<\infty \Leftrightarrow \alpha < 1/2 $$
I'm having some trouble with this. This is what I've done:
\begin{align} \mathbb{E}(T^\alpha) & = \sum_{k=1}^{\infty}k^\alpha\mathbb{P}(T=k) \\ & = \sum_{n=1}^{\infty}(2n)^\alpha \dfrac{1}{2n-1}\binom{2n}{n}2^{-2n} \\&\sim2^\alpha\sqrt{\pi}\sum_{n=0}^{\infty}\dfrac{n^{\alpha-1/2}}{2n-1} \text { (Using Stirlings Approximation) } \end{align}
I would appreciate any help. Thanks!