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Whilst playing around on Wolfram Alpha, I typed in the sum $$\sum_{x=0}^\infty \frac{1}{\binom{2x}{x}}=\frac{2}{27}(18+\pi\sqrt 3)$$ I'm not sure how to derive the answer. My first instinct was to expand the binomial coefficient to get $$\sum_{x=0}^\infty \frac{x!^2}{(2x)!}$$ and then to try using a Taylor Series to get the answer. I thought that if I could find a function $f(n)$ with $$f(n)=\sum_{x=0}^\infty \frac{x!^2n^x}{(2x)!}$$ Then my sum would be equal to $f(1)$. How do I find such a function?

EDIT: I continued on this path and realized that I can use this to set up a recurrence relation for $f^{(x)}(0)$:

$$f^{(0)}(0)=1$$ $$f^{(x)}(0)=\frac{x^2}{2x(2x-1)}f^{(x-1)}(0)$$

However, I'm not sure how this helps me find $f(1)$...

Am I on the right track? Can somebody help me finish what I started, or point me towards a better method of calculating this sum?

Thanks!

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  • $\begingroup$ \begin{eqnarray*} \int_{-1}^{1} (1-x^2)^n dx = \frac{2^{n+1} (n!)^2}{(2n+1)!} \end{eqnarray*} Might help ? $\endgroup$
    – Donald Splutterwit
    Commented Jul 24, 2017 at 15:07
  • $\begingroup$ @DonaldSplutterwit Hmm... I'm not sure how to use that... $\endgroup$
    – Franklin Pezzuti Dyer
    Commented Jul 24, 2017 at 15:09
  • $\begingroup$ Erratum: \begin{eqnarray*} \int_{-1}^{1} (1-x^2)^n dx = \frac{2^{\color{red}{2}n+1} (n!)^2}{(2n+1)!} \end{eqnarray*} $\endgroup$
    – Donald Splutterwit
    Commented Jul 24, 2017 at 17:25
  • $\begingroup$ It's the Maclaurin series of $\left(\frac{\arcsin x}{\sqrt{1-x^{2}}}\right)' $ evaluated at $\frac{1}{2}$. See here and here. $\endgroup$
    – Random Variable
    Commented Jul 26, 2017 at 14:53
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    $\begingroup$ @RandomVariable ...yes, but unless I can just magically recognize any Maclaurin Series at the drop of a hat, I will need to find some other method of finding the answer. Can you tell me how I could find that out without just being able to see it? $\endgroup$
    – Franklin Pezzuti Dyer
    Commented Jul 26, 2017 at 15:03

2 Answers 2

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Hint. One may observe that $$ \frac{1}{\binom{2n}{n}}=n\int_0^1 t^{n-1}(1-t)^ndt,\qquad n\ge1, $$ giving $$ \sum_{n=0}^\infty\frac{1}{\binom{2n}{n}}=1+\int_0^1 \sum_{n=1}^\infty nt^{n-1}(1-t)^n\:dt=1+\int_0^1\frac{t-1}{\left(t^2-t+1\right)^2}dt=\frac{2}{27} \left(18+\sqrt{3} \pi \right) $$ the latter integral is classically evaluated by partial fraction decomposition.

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    $\begingroup$ Nice, thanks for your help! $\endgroup$
    – Franklin Pezzuti Dyer
    Commented Jul 24, 2017 at 15:12
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\begin{eqnarray*} \binom{2n}{n} ^{-1} = \frac{2n+1}{2^{2n+1}} \int_{-1}^{1} (1-x^2)^n dx \end{eqnarray*} Substitute this for summand and inerchange the order of the integral and sum. \begin{eqnarray*} \int_{-1}^{1} \sum_{n=0}^{ \infty} \frac{2n+1}{2^{2n+1}} (1-x^2)^n dx &=& \frac{1}{2} \int_{-1}^{1} \left(2\frac{(\frac{1-x^2}{4})}{(1-(\frac{1-x^2}{4}))^2}+ \frac{1}{1-(\frac{1-x^2}{4})} \right) dx \\ = \int_{-1}^{1} \frac{16}{(3+x^2)^2} dx - \int_{-1}^{1} \frac{2}{(3+x^2)} dx \end{eqnarray*} Now use the standard integrals \begin{eqnarray*} \int_{-1}^{1} \frac{1}{(3+x^2)} dx = \frac{ \pi}{3 \sqrt{3}} \\ \int_{-1}^{1} \frac{1}{(3+x^2)^2} dx = \frac{ 1}{12} + \frac{ \pi}{18 \sqrt{3}} \\ \end{eqnarray*} and the result follows.

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