Euler's work III

Series involving $\zeta(2)$ and Fibonacci numbers

Today we show the proof of this results posted by @seriesbot_q and @infseriesbot \[ 3\sum_{k=1}^{\infty} \frac{1}{k^2\binom{2k}{k}}= \frac{\pi^2}{6} = \zeta(2)\] \[ \sum_{k=1}^{\infty} \frac{F_{2k}}{k^2\binom{2k}{k}} = \frac{4\pi^2}{25\sqrt{5}}\] Both results are particular values of certain series expansion deeply related to the work of Euler. \[ \sum_{k=1}^{\infty} \frac{r^k}{k^2\binom{2k}{k}} = 2\arcsin^2\left(\frac{\sqrt{r}}{2}\right) \] From which we can also conclude \[ \sum_{k=1}^{\infty} \frac{2^k}{k^2\binom{2k}{k}} = \frac{\pi^2}{8} \] \[ \sum_{k=1}^{\infty} \frac{3^k}{k^2\binom{2k}{k}} = \frac{3\pi^2}{9} \]

Proof

First, we will obtain the series expansion for $\displaystyle \arcsin^2(y)$:
\[\arcsin^2(y)= \frac{1}{2}\sum_{k=1}^{\infty} \frac{(k-1)!^2}{(2k)!}(2y)^{2k}=\frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}y^{2k}}{k^2\binom{2k}{k}}\]

This expansion will allow us to find closed-form expressions for the series.

To obtain the series expansion we will rely on the Euler's formula for $\arctan(x)$. We have been using the formula extensively in the past (you can find the proof in one of the propositions of this post) and we need to use it again.

Recall the beautiful Euler's formula: \[\arctan (x) = \frac{x}{1+x^2} \sum_{n=0}^{\infty}\left[\frac{(2n)!!}{(2n+1)!!} \left(\frac{x^2}{1+x^2}\right)^{n}\right]\] Now consider the following identity for inverse trigonometric functions: \[\arcsin(y) = \arctan\left(\frac{y}{\sqrt{1-y^2}}\right)\] For $x= \displaystyle \frac{y}{\sqrt{1-y^2}}$ we have \begin{align*} \arcsin(y) =& \arctan\left(\frac{y}{\sqrt{1-y^2}}\right)\\ =& y\sqrt{1-y^2}\left[\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}y^{2n}\right]\\ =& y\sqrt{1-y^2}\left[1+\frac{2}{3}y^2 +\frac{2}{3}\frac{4}{5}y^4 + \frac{2}{3}\frac{4}{5}\frac{6}{7}y^6...\right]\\ \Longrightarrow \frac{\arcsin(y)}{\sqrt{1-y^2} }=& \sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}y^{2n+1}\\ =& y+\frac{2}{3}y^3+\frac{2}{3}\frac{4}{5}y^5+\frac{2}{3}\frac{4}{5}\frac{6}{7}y^7+... \end{align*} Now consider \begin{align*} \int_{0}^{y} \frac{\arcsin(t)}{\sqrt{1-t^2}dt} =& \int_{0}^{\arcsin(y)} udu \quad (u \mapsto \arcsin(t))\\ =& \left[\frac{u^2}{2}\right]_{0}^{\arcsin(y)} = \frac{\arcsin^2(y)}{2} \end{align*} Therefore \begin{align*} \frac{\arcsin^2(y)}{2} =\int_{0}^{y}\frac{\arcsin(y)}{\sqrt{1-y^2} }dy=& \int_{0}^{y}\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}y^{2n+1}dy\\ & \sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\int_{0}^{y}y^{2n+1}dy\\ & \sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!(2n+2)}y^{2n+2}\\ =& \sum_{k=1}^{\infty} \frac{1}{2k}\frac{(2k-2)!!}{(2k-1)!!}y^{2k} \quad (n=k-1)\\ \end{align*} \begin{align*} \Longrightarrow \arcsin^2(y) =& \sum_{k=1}^{\infty} \frac{1}{k}\frac{(2k-2)!!}{(2k-1)!!}y^{2k}\\ =& \sum_{k=1}^{\infty}\frac{1}{k} \frac{2^{k-1}(k-1)!2^k k!}{(2k)!}y^{2k}\\ =& \frac{1}{2}\sum_{k=1}^{\infty} \frac{(k-1)!^2}{(2k)!}(2y)^{2k} \\ =& \frac{1}{2}\sum_{k=1}^{\infty} \frac{(k)!^2}{k^2(2k)!}2^{2k}y^{2k}\\ =& \frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}y^{2k}}{k^2\binom{2k}{k}} \end{align*} \[\boxed{\therefore\arcsin^2(y)= \frac{1}{2}\sum_{k=1}^{\infty} \frac{(k-1)!^2}{(2k)!}(2y)^{2k}=\frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}y^{2k}}{k^2\binom{2k}{k}}} \] Putting $ \displaystyle y=\frac{1}{2}$ then $\displaystyle \arcsin^2\left(\frac{1}{2}\right)= \frac{\pi^2}{36}$
Then, we have our first result:
\[\boxed{3\sum_{k=1}^{\infty} \frac{(k-1)!^2}{(2k)!} = 3\sum_{k=1}^{\infty} \frac{1}{k^2\binom{2k}{k}}= \frac{\pi^2}{6} = \zeta(2)}\]
Now recall the closed-form expression for the Fibonacci number $\displaystyle F_{j}$:
\[F_{j} = \frac{\phi^{j}-(-\phi)^{-j}}{\sqrt{5}}\]
where $\displaystyle \phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio

Therefore if we put $\displaystyle y = \frac{\sqrt{5}+1}{4} \Longrightarrow \arcsin^2\left(\frac{\sqrt{5}+1}{4} \right) =\frac{9\pi^2}{100}$
Hence \[ \frac{9\pi^2}{100}= \frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}\left(\frac{\sqrt{5}+1}{4}\right)^{2k}}{k^2\binom{2k}{k}} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{\phi^{2k}}{k^2\binom{2k}{k}} \] Now if we put $\displaystyle y = -\frac{1}{\sqrt{5}+1} \Longrightarrow \arcsin^2\left( -\frac{1}{\sqrt{5}+1}\right) =\frac{\pi^2}{100}$
Hence \[ \frac{\pi^2}{100}= \frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}\left( -\frac{1}{\sqrt{5}+1}\right)^{2k}}{k^2\binom{2k}{k}} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{(-\phi)^{-2k}}{k^2\binom{2k}{k}} \]
Therefore \[ \sum_{k=1}^{\infty} \frac{\phi^{2k}}{k^2\binom{2k}{k}}- \sum_{k=1}^{\infty} \frac{(-\phi)^{-2k}}{k^2\binom{2k}{k}} = \sum_{k=1}^{\infty} \frac{\phi^{2k}- (-\phi)^{-2k}}{k^2\binom{2k}{k}} = \frac{8\pi^2}{50} = \frac{4\pi^2}{25}\] Then \[ \frac{1}{\sqrt{5}}\sum_{k=1}^{\infty} \frac{\phi^{2k}- (-\phi)^{-2k}}{k^2\binom{2k}{k}} = \frac{4\pi^2}{25\sqrt{5}}\]
Hence, we can conclude \[\boxed{ \sum_{k=1}^{\infty} \frac{F_{2k}}{k^2\binom{2k}{k}} = \frac{4\pi^2}{25\sqrt{5}}}\]