# I have come across the following series, which I suspect converges to ln2: sum_(k=1)^oo (1)(4^k(2k))((2k),(k)) I could not derive this series from some of the standard expressions for ln2. The sum of the first 100000 terms agrees with ln2 only up to two digits.

A series converging (or not) to $\mathrm{ln}2$
I have come across the following series, which I suspect converges to $\mathrm{ln}2$
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{4}^{k}\left(2k\right)}\left(\genfrac{}{}{0}{}{2k}{k}\right).$
I could not derive this series from some of the standard expressions for $\mathrm{ln}2$. The sum of the first
$100000$ terms agrees with $\mathrm{ln}2$ only up to two digits.
Does the series converge to $\mathrm{ln}2$?
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Abbigail Vaughn
Hint: (or outline -- a lot of details and justifications must be done where there are $\left(\star \right)$'s)
for $|x|<\frac{1}{4}$,
$\begin{array}{rl}f\left(x\right)& \stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{2k}{k}\right)\frac{{x}^{k}}{k}=\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{2k}{k}\right){\int }_{0}^{x}{t}^{k-1}dt\\ & \stackrel{\left(\star \right)}{=}\frac{1}{2}{\int }_{0}^{x}\left(\sum _{k=1}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{2k}{k}\right){t}^{k-1}\right)dt=\frac{1}{2}{\int }_{0}^{x}\frac{1}{t}\left(\sum _{k=1}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{2k}{k}\right){t}^{k}\right)dt\\ & \stackrel{\left(\star \right)}{=}\frac{1}{2}{\int }_{0}^{x}\frac{1}{t}\left(\frac{1}{\sqrt{1-4t}}-1\right)dt\\ & =\frac{1}{2}2\mathrm{ln}\frac{2}{\sqrt{1-4x}+1}\end{array}$
"so" ($\star$)
$\begin{array}{}\text{(}\star \text{)}& f\left(x\right)\underset{t\to {\frac{1}{4}}^{-}}{\overset{}{\to }}\mathrm{ln}2\end{array}$
###### Did you like this example?
ganolrifv9
Recall that for $|x|<1$ we have
$\frac{1}{\sqrt{1-{x}^{2}}}=\sum _{k=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{2n}{n}\right)\frac{{x}^{2n}}{{4}^{n}}$
shifting over the first terms on the series and dividing by x gives
$\sum _{n=1}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{2n}{n}\right)\frac{{x}^{2n-1}}{{4}^{n}}=\frac{1-\sqrt{1-{x}^{2}}}{\sqrt{1-{x}^{2}}}\frac{1}{x}$
Now given an small $\epsilon >0$ we can integrate both sides from $0$ to $1-\epsilon$, the swap on the RHS is justified due to the uniform convergence, next we let $\epsilon$ shrink to zero. Wallis product formula yields
$\frac{1}{{4}^{n}2n}\left(\genfrac{}{}{0}{}{2n}{n}\right)\sim \frac{1}{\sqrt{2\pi }}\frac{1}{n\sqrt{2n+1}}$
Now given an small $\epsilon >0$ we can integrate both sides from $0$ to $1-\epsilon$, the swap on the RHS is justified due to the uniform convergence, next we let $\epsilon$ shrink to zero. Wallis product formula yields
$\frac{1}{{4}^{n}2n}\left(\genfrac{}{}{0}{}{2n}{n}\right)\sim \frac{1}{\sqrt{2\pi }}\frac{1}{n\sqrt{2n+1}}$
which allows us to move in the limit inside and letting $\epsilon \to 0$ Hence we have that
$\sum _{n=1}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{2n}{n}\right)\frac{1}{{4}^{n}2n}={\int }_{0}^{1}\frac{1-\sqrt{1-{x}^{2}}}{\sqrt{1-{x}^{2}}}\frac{dx}{x}={\int }_{0}^{\pi /2}\frac{1-\mathrm{cos}\left(u\right)}{\mathrm{sin}\left(u\right)}du={\int }_{0}^{\pi /2}\frac{\mathrm{sin}\left(u/2\right)}{\mathrm{cos}\left(u/2\right)}du$
It is pretty obvious what to do next..