Euler's work V

Consequences of the series expansion of $\displaystyle \frac{1}{\sqrt{1-x}}$ vol II

We continue developing some results of series involving the binomial coefficient. This time we show the proof of the following beautiful result posted by @seriesbot_q:
\[\sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2} = \frac{\sqrt{3}}{2}\left(1+\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2} + \frac{1}{7^2}+\frac{1}{8^2}-...\right) \]

Proof:

From the basic series \[\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n} }x^n = \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!}x^{n} \] If we make the transformation $\displaystyle x \mapsto x^2$ \[\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n} }x^{2n} \] Integrating from $0$ to $t$: \[\int_{0}^{t} \frac{1}{\sqrt{1-x^2}} dx = \arcsin(t) = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n+1) }t^{2n+1} \] Dividing by $t$: \[\frac{\arcsin(t)}{t} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n+1) }t^{2n} \] Integrating again from $0$ to $\displaystyle \frac{1}{2}$: \[\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n+1) }\int_{0}^{\frac{1}{2}}t^{2n}dt = \frac{1}{2}\sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2 } \] \[ \Longrightarrow \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2} = 2\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt \tag{1}\]
The left hand side is the series that we are looking for. Despite the fact that the integral $\displaystyle \int \frac{\arcsin x}{x}dx$ cannot be expressed as a finite combination of elementary functions, we can express it in the form of a new infinite series.

First we have to make a change of variable: \[ \int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt = \int_{0}^{\frac{\pi}{6}} \frac{w\cos(w)}{\sin(w)} dw = \int_{0}^{\frac{\pi}{6}} w\cot(w) dw\quad (w\mapsto \arcsin(t))\] Now, recall the following series: \[ \cot(w) = \lim_{t \to 0}2\sum_{k=1}^{\infty} e^{-tk}\sin(2kw)\] This proof of this result is easy: Consider the "discrete" Laplace transform: \[\sum_{k=1}^{\infty} e^{-kt}\sin(kw) = \frac{1}{2} \frac{\sin w}{\cosh t - \cos w} \quad t>0\] This follows from: \begin{align*} \sum_{k=1}^{\infty} e^{-kt}\sin(kw) =& \sum_{k=0}^{\infty} e^{-kt}\left(\frac{e^{kiw}-e^{-kiw}}{2i}\right)\\ =& \frac{1}{2i} \sum_{k=0}^{\infty} e^{-kt}\left(e^{kiw}-e^{-kiw}\right)\\ =& \frac{1}{2i} \sum_{k=0}^{\infty} e^{k(iw-t)} -\frac{1}{2} \sum_{k=0}^{\infty} e^{k(-t-iw)}\\ =& \frac{1}{2i} \frac{1}{1-e^{iw-t}}- \frac{1}{2i} \frac{1}{1-e^{-t-iw}} \\ =& \frac{1}{2i} \left(\frac{e^t}{e^t-e^{iw}}- \frac{e^t}{e^t-e^{-iw}} \right)\\ =& \frac{1}{2i} \left(\frac{e^{t+iw}-e^{t-iw}}{e^{2t}-e^{t-iw}-e^{t+iw}+1}\right)\\ =& \frac{1}{2i} \left(\frac{e^{iw}-e^{-iw}}{e^{t}+e^{-t}-(e^{-iw}+e^{iw})}\right)\\ =& \frac{1}{2} \frac{\frac{e^{iw}-e^{-iw}}{2i}}{\frac{e^{t}+e^{-t}-(e^{-iw}+e^{iw})}{2}}\\ =& \frac{1}{2} \frac{\sin w}{\cosh t - \cos w}\\ \end{align*} If we take the limit as $\displaystyle t \to 0+$ \[\lim_{t \to 0+ }\sum_{k=1}^{\infty}e^{tk} \sin(kw) = \frac{1}{2} \frac{\sin w}{1 - \cos w} \] and using the following identities \[ \sin\left(\theta\right) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\] \[1-\cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right)\] we have \[ \lim_{t \to 0+} \sum_{k=1}^{\infty} e^{-tk}\sin(kw) = \frac{1}{2} \frac{\sin w}{1 - \cos w} = \frac{1}{2}\cot\left(\frac{w}{2}\right)\] therefore \[ \cot\left(w\right) = \lim_{t \to 0+} 2\sum_{k=1}^{\infty} e^{-tk}\sin(2kw) \tag{2} \] This seems to imply that \[ \cot\left(w\right) \stackrel{?}{=} 2\sum_{k=1}^{\infty} \sin(2kw) \]
However, this is not true... at least formally: note that the series $\displaystyle \sum_{k=1}^{ \infty}\sin(2kw)$ does not converge.

But we can calculate the integral taking the limit as $t \to 0+$: \begin{align*} I = \int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt =& \int_{0}^{\frac{\pi}{6}} \frac{w\cos(w)}{\sin(w)} dw\\ =& \int_{0}^{\frac{\pi}{6}} w\cot(w) dw\\ =& 2\int_{0}^{\frac{\pi}{6}} w\left[\lim_{t \to 0+} \sum_{k=1}^{\infty}e^{-tk}\sin(2kw)\right] dw \quad (\textrm{from (2)})\\ =& \lim_{t \to 0+} 2\sum_{k=1}^{\infty} e^{-tk}\int_{0}^{\frac{\pi}{6}} w\sin(2kw) dw\\ =& \lim_{t \to 0+} \frac{1}{2}\sum_{k=1}^{\infty}e^{-tk} \frac{1}{k^2}\int_{0}^{\frac{\pi k}{3}} s\sin(s) dw \quad (s \mapsto 2kw)\\ \stackrel{IBP}{=}& \lim_{t \to 0+} \sum_{k=1}^{\infty}e^{-tk}\frac{1}{2k^2} \left[-\frac{\pi k}{3}\cos\left(\frac{k\pi}{3}\right) + \int_{0}^{\frac{\pi k}{3}} \cos sds \right]\\ =& \lim_{t \to 0+} \sum_{k=1}^{\infty}e^{-tk}\frac{1}{2k^2} \left[\sin\left(\frac{\pi k}{3}\right)-\frac{\pi k}{3}\cos\left(\frac{k\pi}{3}\right) \right]\\ =& \lim_{t \to 0+} \frac{1}{2} \sum_{k=1}^{\infty} e^{-tk}\frac{\sin\left(\frac{\pi k}{3}\right)}{k^2}-\lim_{t \to 0} \frac{\pi}{6}\sum_{k=1}^{\infty}e^{-tk}\frac{\cos\left(\frac{k\pi}{3}\right)}{k}\\ =& \frac{1}{2} \sum_{k=1}^{\infty}\frac{\sin\left(\frac{\pi k}{3}\right)}{k^2}- \frac{\pi}{6}\sum_{k=1}^{\infty}\frac{\cos\left(\frac{k\pi}{3}\right)}{k}\\ \end{align*} where the last equality follows from the fact that the series \[ \sum_{k=1}^{\infty}\frac{\sin\left(\frac{\pi k}{3}\right)}{k^2} \quad \sum_{k=1}^{\infty}\frac{\cos\left(\frac{k\pi}{3}\right)}{k}\] are well known Fourier series that converge.

Now we will prove that the second series is equal to zero:

From Euler's formula we have \[\sum_{k=1}^{\infty} \frac{\cos\left(\frac{k\pi}{3}\right)}{k} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{e^{\frac{i\pi k}{3}}}{k}+ \frac{1}{2}\sum_{k=1}^{\infty} \frac{e^{-\frac{i\pi k}{3}}}{k}\] We can split the sum into two because the series \[ \sum_{k=1}^{\infty} \frac{z^k}{k}\] converges when $\displaystyle |z|<1$. But this series is just the Laurent series for the principal branch of the $-\ln(1-z)$ function: \[ \ln(1-z) = -\sum_{k=1}^{\infty} \frac{z^k}{k} \] Hence we can write: \[\sum_{k=1}^{\infty} \frac{\cos\left(\frac{k\pi}{3}\right)}{k} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{e^{\frac{i\pi k}{3}}}{k}+ \frac{1}{2}\sum_{k=1}^{\infty} \frac{e^{-\frac{i\pi k}{3}}}{k} = -\frac{1}{2}\ln\left(1-e^{\frac{i\pi}{3}}\right)-\frac{1}{2}\ln\left(1-e^{-\frac{i\pi}{3}}\right) = -i\frac{\pi}{6}+ i\frac{\pi}{6} =0\] Therefore \[I =\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt = \frac{1}{2} \sum_{k=1}^{\infty} \frac{\sin\left(\frac{\pi k}{3}\right)}{k^2} \] If we tabulate the sequence $\displaystyle \sin\left(\frac{\pi k}{3}\right) \textrm{ for } k=1,2,3,...$ we find: \[ \left(\sin\left(\frac{\pi k}{3}\right)\right)_{k} = \left(\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 0 ,-\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2},0, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2},... \right) = \frac{\sqrt{3}}{2}(1,1,0,-1-1,0,1,1,...)\] Therefore \[I =\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt = \frac{1}{2} \sum_{k=1}^{\infty} \frac{\sin\left(\frac{\pi k}{3}\right)}{k^2}= \frac{\sqrt{3}}{4}\left(1+\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2} + \frac{1}{7^2}+\frac{1}{8^2}+...\right) \tag{3} \] Putting all the ingredients together: \begin{align*} \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2} = & 2\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt \quad (\textrm{from (1)}) \\ =& \frac{\sqrt{3}}{2}\left(1+\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2} + \frac{1}{7^2}+\frac{1}{8^2}+...\right) \quad (\textrm{from (3)}) \end{align*} Therefore \[\boxed{\sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2} = \frac{\sqrt{3}}{2}\left(1+\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2} + \frac{1}{7^2}+\frac{1}{8^2}-...\right) }\]