While reading slides to one of my courses I came across the following: $\sum_{i=0}^\infty {2i \choose i} a^i$. The original series was a little more complicated but is essentially equivalent to this one. In the slides, the professor had written that the series is equal to $\frac{1}{\sqrt{1 - 4a}}$.
Could you please provide me a hint on how to approach this problem? When I used Stirling's approximation I was able to obtain $1 + \sum_{i=1}^\infty \frac{1}{\sqrt{\pi i}} a^i$, is this the right way? My guess is not, since in the slides there is an equal sign, not an approximation.
