TITLE: A series converging (or not) to $\ln 2$
QUESTION [3 upvotes]: I have come across the following series, which I suspect converges to $\ln 2$:
$$\sum_{k=1}^\infty \frac{1}{4^k(2k)}\binom{2k}{k}.$$
I could not derive this series from some of the standard expressions for $\ln 2$.  The sum of the first $100 000$ terms agrees with $\ln 2$ only up to two digits.  
Does the series converge to $\ln 2$?

REPLY [0 votes]: Recall that for $|x|<1$ we have $$\frac{1}{\sqrt{1-x^{2}}}=\sum_{k=0}^{\infty}{2n \choose n}\frac{x^{2n}}{4^{n}}$$ shifting over the first terms on the series and dividing by x gives $$\sum_{n=1}^{\infty}{2n\choose n}\frac{x^{2n-1}}{4^{n}}= \frac{1-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\frac{1}{x}$$ Now given an small $\varepsilon>0$ we can integrate both sides from $0$ to $1-\varepsilon$, the swap on the RHS is justified due to the uniform convergence, next we let $\varepsilon$ shrink to zero. Wallis product formula yields $$\frac{1}{4^{n}2n}{2n \choose n}\sim\frac{1}{\sqrt{2\pi}}\frac{1}{n\sqrt{2n+1}}$$ which allows us to move in the limit inside and letting $\varepsilon \rightarrow 0$ Hence we have that
$$\sum_{n=1}^{\infty}{2n\choose n}\frac{1}{4^{n}2n}= \int_{0}^{1}{\frac{1-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\frac{dx}{x}}=\int_{0}^{\pi/2}\frac{1-\cos(u)}{\sin(u)}du=\int_{0}^{\pi/2}\frac{\sin(u/2)}{\cos(u/2)}du$$
It is pretty obvious what to do next..
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