I've encountered the following series:
$$\sum_{t=1}^\infty {1 \over 2^{t}}\, {{\large t} \choose {\large{t + x \over 2}}}$$
Is this series even convergent? I'm really lacking knowledge on series that contains binomial coefficients.
I actually encountered this series when calculating $E(T_x)$, where $T_x$ is the first time a random walker hits the disposition $x$. I'm also looking for other ways to get around this series via manipulation on expected value, but no clue yet..
Many thanks!
This diverges for all $x$. For $x=0$, Stirling's approximation gives us $$2^{-t}\binom{t}{t/2}=\frac{t!}{2^t((t/2)!)^2}\sim \frac{\sqrt{2\pi t}t^te^{2\cdot t/2}}{2^t\cdot 2\pi (t/2)(t/2)^{2\cdot t/2}e^t}=\sqrt{2\pi/t}$$ and $\sum_{t=1}^\infty \sqrt{2\pi/t}=\infty$. Since $$P_x(t):=\frac{t+x}{2}!\frac{t-x}{2}!-((t/2)!)^2$$ is a polynomial in $t$, we have $$\binom{t}{\frac{t+x}{2}}=\frac{t!}{\frac{t+x}{2}!\frac{t-x}{2}!}=\frac{t!}{((t/2)!)^2+P_x(t)}\sim \frac{t!}{((t/2)!)^2}=\binom{t}{t/2}$$ and so similarly the series diverges for all values of $x$.