<s>[INST] A series converging (or not) to $\ln 2$ 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 $\ln 2$? 
 [/INST] Hint: (or outline -- a lot of details and justifications must be done where there are $(\star)$'s)


*

*for $\lvert x\rvert < \frac{1}{4}$, 
\begin{align*}
f(x)&\stackrel{\rm def}{=}\frac{1}{2}\sum_{k=1}^{\infty} \binom{2k}{k}\frac{x^k}{k} = \frac{1}{2}\sum_{k=1}^{\infty} \binom{2k}{k}\int_0^x t^{k-1}dt \\
&\stackrel{(\star)}{=} \frac{1}{2} \int_0^x \left(\sum_{k=1}^{\infty} \binom{2k}{k} t^{k-1}\right)dt = \frac{1}{2} \int_0^x \frac{1}{t} \left(\sum_{k=1}^{\infty} \binom{2k}{k} t^{k}\right)dt \\
&\stackrel{(\star)}{=} \frac{1}{2} \int_0^x \frac{1}{t} \left(\frac{1}{\sqrt{1-4t}}-1\right)dt \\
&= \frac{1}{2} 2\ln\frac{2}{\sqrt{1-4x}+1}
\end{align*}

*"so" ($\star$)
\begin{align*}
f(x)\xrightarrow[t\to\frac{1}{4}^-]{}\ln 2 \tag{$\star$}
\end{align*}

</s>