Closed form for this particular Taylor series

Question

I know that in general one can't find closed forms for arbitrary infinite series, but in working on a problem I came across this sum:

$$\sum_{n=0}^\infty \binom{2n}{n}(1/9)^{n}$$

(Note: I originally put $(1/3)^{n}$, have corrected it above.)

The Taylor series

$$\sum_{n=0}^\infty \binom{2n}{n} x^{n}$$

looked vaguely familiar and so I tried to find/derive a closed form for this but have had no luck. So my questions are:

• Do you recognize this particular sum?

• Do you have any suggestions for tracking down questions like this? I checked most of the basic Calc I level functions, and did an initial scan through Abromowitz and Stegun, but didn't find anything even close.

• Are there any methods that might apply? I have a memory of a paper where the author had a method for a wide variety of sums like these with binomial co-efficients, but I can't track it down.

Note: From the "Related Questions" in the side bar here I'm going to check if hypergeometric series can help me, but I'll go ahead and post this anyways.

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Edit: Thanks for the identification. I'd like to know if it's just something you've recognized, or you know how to search Wolfram better than I do. For example, I wouldn't know to call the co-efficient the "Central Binomial".

Answer 1

In order to find the generating function for central binomial coefficients, one may notice that $\int_{-\pi}^{\pi}e^{ni\theta}\,d\theta = 2\pi \delta(n)$ implies $$ \int_{-\pi}^{\pi}\cos(\theta)^{2n}\,d\theta = \int_{-\pi}^{\pi}\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^{2n}\,d\theta =\frac{2\pi}{4^n}\binom{2n}{n}\tag{1}$$ hence: $$ S=\sum_{n\geq 0}\binom{2n}{n}\frac{1}{9^n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{n\geq 0}\left(\frac{2}{3}\cos\theta\right)^{2n}\,d\theta = \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{d\theta}{1-\frac{4}{9}\cos^2\theta}\tag{2}$$ and the problem boils down to computing the last integral. We have: $$ S = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1-\frac{4}{9}\cos^2\theta}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{dt}{1+t^2-\frac{4}{9}}\tag{3}$$ through the substitution $\theta=\arctan t$, and by setting $t=u\sqrt{\frac{5}{9}}$ the identity $$ S = \color{red}{\frac{3}{\sqrt{5}}}\tag{4}$$ easily follows. A self-contained proof.

Answer 2

We have that for $|x|< 1/4$, $$\sum_{n=0}^\infty \binom{2n}{n} x^{n}=\sum_{n=0}^\infty \binom{-1/2}{n} (-4x)^{n}=\frac{1}{\sqrt{1-4x}}.$$ see here: http://mathworld.wolfram.com/CentralBinomialCoefficient.html

The convergence for $|x|< 1/4$ follows from the fact that, by Stirling approximation, $$\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}.$$

Since $|1/9|<1/4$, we obtain $$\sum_{n=0}^\infty \binom{2n}{n}(1/9)^{n}=\frac{1}{\sqrt{1-4/9}}=\frac{3}{\sqrt{5}}.$$