## Alternating Sum of the Reciprocals of the Central Binomial Coefficients

In the last post we proved the generating function for the reciprocals of the central binomial coefficients: $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{\binom{2n}{n}} = \frac{1}{1-\frac{x}{4}} + \frac{4 \sqrt{x}\arcsin \left(\frac{\sqrt{x}}{2}\right)}{(4-x)^{3/2}}.$

In this post we’re going to use this generating function to find the alternating sum of the reciprocals of the central binomial coefficients.  On the left side of the generating function, we need merely substitute $-1$ for x.  (Recall that the series converges for $-4 < x < 4$.)  On the right side, however, we have the arcsine of an imaginary number.  Most of this post will be about how to make sense of that.  Essentially, we’re going to convert inverse sine to inverse hyperbolic sine and then use the logarithm formula for the latter.

First, recall the representation of $\sin x$ in terms of complex exponentials, as well as the definition of hyperbolic sine: $\displaystyle \sin x = \frac{e^{ix} - e^{-ix}}{2i}, \\ \sinh x = \frac{e^x - e^{-x}}{2}$.

Substituting ix for x in the representation of $\sin x$ shows that $\sin (ix) = i \sinh x$.  Now, letting $y = \sinh x$, we have $x = \sinh^{-1} y$.  Also, $\sin (ix) = i y$.  Thus $ix = \arcsin (iy)$, and $i \sinh^{-1} y = \arcsin (iy)$.  In other words, $\arcsin (ix) = i \sinh^{-1} x$.

With the representation $\displaystyle \sinh^{-1} x = \ln (x + \sqrt{x^2+1})$,

we can finally obtain a simple expression for the alternating sum of the reciprocals of the central binomial coefficients: $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{\binom{2n}{n}} = \frac{1}{1-\frac{-1}{4}} + \frac{4 (\sqrt{-1})\arcsin \left(\frac{\sqrt{-1}}{2}\right)}{(4-(-1))^{3/2}} = \frac{4}{5} + \frac{4 i \arcsin \left(\frac{i}{2}\right)}{5^{3/2}}$ $\displaystyle = \frac{4}{5} + \frac{4 i (i) \sinh^{-1} \left(\frac{1}{2}\right)}{5 \sqrt{5}} = \frac{4}{5} - \frac{4 \ln (\frac{1}{2} + \sqrt{1/4+1})}{5 \sqrt{5}} =\frac{4}{5} - \frac{4 \sqrt{5}}{25} \ln \left(\frac{1 + \sqrt{5}}{2} \right).$

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