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Evaluating the function $S(a) = \sum_{n=0}^{\infty} a^n \binom{2n}{n}$.

The first thing that I noticed is that $S(a)$ converges when $a \in [-\frac{1}{4}, \frac{1}{4})$ because $$ \binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}. $$

However, this does not help me actually find $S(a)$. Furthermore, various strategies such as telescoping and DuSS (differentiation under the summation sign) have been foiled by the pesky binomial coefficient. If we switch to the Gamma function, DuSS is possible but extremely messy. I'm wondering what approach I should utilize in order to find an expression for $S(a)$ in terms of elementary functions.

Jendrik Stelzner
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Display name
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  • Do you expect a closed form to exist? – Arthur Aug 27 '18 at 06:03
  • @Arthur An inverse calculator gave the guess $S(-1/8) = \sqrt{2/3},$ so I'm confident that there's a general form for $S(a).$ The appearance of the square root suggests multiplying the series by itself and trying various double summation tactics, but this route has not led anywhere. – Display name Aug 27 '18 at 06:08
  • We have $\sum 1/n^2=\pi^2/6$, but just because we know its value at $2$, and it looks nice, doesn't mean that we have a closed form expression for the Riemann zeta function. – Arthur Aug 27 '18 at 06:25

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A well-known formula is $$\sum_{n=0}^\infty\binom{2n}{n}x^n=\frac1{\sqrt{1-4x} }$$ for $|x|<1/4$. This can be verified by expanding the RHS via the binomial theorem.

Angina Seng
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  • I saw this formula before but had forgotten it by now. Now I'm disappointed because the problem I posted was intended by whoever placed it to be a difficult math competition question, but turned out to be a "trivial by memorized identity; you either know it or you don't" sort of problem. – Display name Aug 27 '18 at 06:20