Find the value of the series

Question

I have the series $$\sum_{n=0}^{\infty} \binom{2n}{n} \frac{1}{7^{n}}$$ My hint to find the value is to use Cauchy's integral formula. I know the function to consider is $f(z)=(1+z)^{n}$, using Newton's binomial then $$(1+z)^{n}=\sum_{k=0}^{n}\binom{n}{k}z^{k}$$ From this point, I don't know how to continue.

Answer 1

Consider the generating functions $$P(n)=\sum_{n=0}^\infty\binom{2n}{n}x^n\text{ and } C(n)=\sum_{n=0}^\infty C_nx^n$$ where $C_n=\frac{1}{n+1}\binom{2n}{n}$ denotes the nth Catalan number.

Recall that $C_n$ counts the amount of ways to go from $(0,0)$ to $(n,n)$, moving right and up, without going above the line $y=x$. For fixed $n$, define $x_i$, $0\le i<n$ to be the number of Catalan paths where the last point at which the path touches $y=x$ is at $(i,i)$. Then, it is clear that $x_i=C_i\cdot C_{n-i-1}$, so we have the recurrence $$C_n=\sum_{i=0}^{n-1}C_i\cdot C_{n-i-1}$$ Thus, it's generating function must satisfy the equation $$C(x)=1+xC(x)^2\implies C(x)=\frac{1\pm\sqrt{1-4x}}{2x}$$ $x\to0$ should give $C(x)=1$, so we take the negative. From the closed form for $C_n$, we have that $P(n)=\frac{d}{dx}\left(xC(x)\right)$, so $$P(n)=\frac{d}{dx}\frac{1-\sqrt{1-4x}}{2}=\frac{d}{dx}\frac{-\sqrt{1-4x}}{2}=\frac{1}{\sqrt{1-4x}}$$ Hence, the answer to the initial question is $P(1/7)=\sqrt{7/3}$.

Answer 2

Disclaimer: This does not use any complex analysis so it might not be the solution you're looking for.

My first thought is to use generating functions. We have a $(1/7)^n$ term in the series, which reminds us that we can probably use the function $f(x):=\sum_{n\geq0}\binom{2n}{n}x^n$, find an explicit expression for $f(x)$, and then just substitute $x=1/7$ to solve the problem. From this question we see indeed that $$f(x)=\sum_{n\geq0}\binom{2n}{n}x^n=\frac1{\sqrt{1-4x}},$$

so the answer will simply be $f(1/7)=\sqrt{7/3}$.