How can I evaluate $\sum_{i=0}^\infty \frac{1}{k^i} \binom{2i}{i}$ Evaluate $$\sum_{i=0}^\infty \left(\frac{\binom{2i}{i}}{k^i}\right),$$
where $k$ is a whole number.
I can't figure out how to approach this question, as no binomial series has such coefficients.
 A: Note, that
$$\binom{2i}{i}=(-4)^i\binom{-\frac{1}{2}}{i}$$

So, we can write OPs series as binomial series
\begin{align*}
  \sum_{i=0}^{\infty}\binom{2i}{i}\frac{1}{k^i}
  &=\sum_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}\left(-\frac{4}{k}\right)^i\\
  &=\frac{1}{\sqrt{1-\frac{4}{k}}}\\
  &=\sqrt{\frac{k}{k-4}}
  \end{align*}
convergent for $\left|-\frac{4}{k}\right|<1$, i.e. $k>4$.


[2016-01-14] Addendum

We can extend the binomial coefficient for arbitrary $\alpha\in\mathbb{C}$ and $n\in\mathbb{N}$
  \begin{align*}
\binom{\alpha}{n}:=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n(n-1)\cdots3\cdot2\cdot1}
\end{align*}
  Putting $\alpha=-\frac{1}{2}$ we obtain
  \begin{align*}
\binom{-\frac{1}{2}}{n}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\\
&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)}{n!}\\
&=\left(-\frac{1}{2}\right)^n\frac{1}{n!}(2n-1)!!\tag{1}\\
&=\left(-\frac{1}{2}\right)^n\frac{1}{n!}\cdot\frac{(2n)!}{(2n)!!}\\
&=\left(-\frac{1}{2}\right)^n\frac{1}{n!}\cdot\frac{(2n)!}{n!2^n}\tag{2}\\
&=\left(-\frac{1}{4}\right)^n\frac{(2n)!}{(n!)^2}\\
&=\left(-\frac{1}{4}\right)^n\binom{2n}{n}
\end{align*}

Comment:


*

*In (1) we use double factorials and the relation $(2n)!=(2n)!!(2n-1)!!$

*In (2) we use $(2n)!!=(2n)(2n-2)\cdots4\cdot2=n!2^n$
