Sum of the Reciprocals of the Central Binomial Coefficients

In this post we prove the formula for the sum of the reciprocals of the central binomial coefficients \binom{2n}{n}:

\displaystyle \sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}} = \frac{4}{3}  + \frac{2 \pi \sqrt{3}}{27}.

(Of course, the sum of the central binomial coefficients themselves does not converge.)

We start with the beta integral\displaystyle \frac{1}{\binom{n}{k}} = (n+1) \int_0^1 x^k (1-x)^{n-k} \, dx.

Replacing n with 2n and k with n, this becomes

\displaystyle \frac{1}{\binom{2n}{n}} = (2n+1) \int_0^1 x^n (1-x)^n \, dx = \int_0^1 (2n+1) (x(1-x))^n \, dx.

Now, let y = \sqrt{x(1-x)} and sum both sides from 0 to \infty, moving the summation inside the integral.  (We’re not performing a u-substitution here; the bounds and the differential remain in terms of x.  The substitution is just to make the upcoming infinite series calculation easier to see.)  We have

\displaystyle \sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}} = \int_0^1 \sum_{n=0}^{\infty} (2n+1) y^{2n} \, dx = \int_0^1 \sum_{n=0}^{\infty}  \frac{d}{dy} (y^{2n+1}) \, dx = \int_0^1 \frac{d}{dy} \sum_{n=0}^{\infty}   y^{2n+1} \, dx

\displaystyle = \int_0^1  \frac{d}{dy} \left( y \sum_{n=0}^{\infty}  (y^2)^n \right) \, dx = \int_0^1 \frac{d}{dy} \left( \frac{y}{1-y^2} \right) \, dx = \int_0^1 \frac{1+y^2}{(1-y^2)^2} dx

\displaystyle = \int_0^1 \frac{1+x-x^2}{(1-x+x^2)^2} dx.

(The infinite series \sum_{n=0}^{\infty} (y^2)^n converges because x must be between 0 and 1, which means the maximum value of y = \sqrt{x(1-x)} is 1/2.)

At this point we have the integral of a rational function.  Partial fractions decomposition (or just rewriting the numerator as 2 - (1-x+x^2)) gets us to

\displaystyle \int_0^1 \left(\frac{-1}{(1-x+x^2)^2} + \frac{2}{(1-x+x^2)^2}\right)  dx.

Using trigonometric substitution, this evaluates to

\displaystyle \left. \frac{2\sqrt{3}}{9} \arctan \left( \frac{2x-1}{\sqrt{3}} \right) + \frac{2(2x-1)}{3(1-x+x^2)} \right|_0^1

\displaystyle = \frac{2\sqrt{3}}{9} \arctan \frac{1}{\sqrt{3}} + \frac{2}{3} - \frac{2 \sqrt{3}}{9} \arctan \left(\frac{-1}{\sqrt{3}}\right) - \frac{-2}{3}

\displaystyle = \frac{4}{3} + \frac{2\sqrt{3}}{9} \left( \frac{\pi}{6} - \frac{-\pi}{6} \right) = \frac{4}{3} + \frac{2\pi \sqrt{3}}{27}.

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One Response to Sum of the Reciprocals of the Central Binomial Coefficients

  1. Pingback: Generating Function for the Reciprocals of the Central Binomial Coefficients | A Narrow Margin

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