How can I evaluate $\sum_{i=0}^\infty \frac{1}{k^i} \binom{2i}{i}$

Question

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.

Answer 1

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. $k>4$.

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[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 $(2n)!!=(2n)(2n-2)\cdots4\cdot2=n!2^n$