2

Wikipedia's "List of formulae involving $\pi$" entry states that $$\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}.$$ Why is this true?

If $\varphi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2$, then we can rewrite this identity as $$F(\varphi)-F(\psi)=\frac{4\pi^2}{25}\quad\text{ where }F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}.$$ $F(x)$ looks similar to the power series for arcsin.

Blue
  • 75,673
Thomas Browning
  • 4,351

2 Answers2

7

$$F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}=2 \left[\sin ^{-1}\left(\frac{x}{2}\right)\right]^2$$ makes $$F(\varphi)=\frac{9 \pi ^2}{50}\qquad \text{and}\qquad F(\psi)=\frac{\pi ^2}{50}$$ since $$\varphi=1-2 \cos \left(\frac{3 \pi }{5}\right)$$

Claude Leibovici
  • 260,315
3

In this answer, it is shown that $$ \frac{\arcsin(x)}{\sqrt{1-x^2}} =\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}\tag1 $$ Integrating $(1)$ and multiplying by $4$ yields $$ \begin{align} 2\arcsin^2(x) &=\sum_{k=0}^\infty\frac1{(2k+1)(2k+2)}\frac{4^{k+1}}{\binom{2k}{k}}x^{2k+2}\\ &=\sum_{k=0}^\infty\frac{4^{k+1}x^{2k+2}}{(k+1)^2\binom{2k+2}{k+1}}\\ &=\sum_{k=1}^\infty\frac{4^kx^{2k}}{k^2\binom{2k}{k}}\tag2 \end{align} $$ Applying $F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}$, we get $$ \begin{align} \sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}} &=\frac2{\sqrt5}\left(\arcsin^2\left(\frac\phi2\right)-\arcsin^2\left(\frac{1-\phi}2\right)\right)\\ &=\frac2{\sqrt5}\left(\frac{9\pi^2}{100}-\frac{\pi^2}{100}\right)\\ &=\frac{4\pi^2}{25\sqrt5}\tag3 \end{align} $$

Addendum: Trigonometric Values Used Above

Note that $$ \begin{align} 4\cos^2(2\pi/5)+2\cos(2\pi/5)-1 &=\left(e^{2\pi i/5}+e^{-2\pi i/5}\right)^2+\left(e^{2\pi i/5}+e^{-2\pi i/5}\right)-1\\[6pt] &=e^{4\pi i/5}+e^{2\pi i/5}+1+e^{-2\pi i/5}+e^{-4\pi i/5}\\ &=\frac{e^{10\pi i/5}-1}{e^{2\pi i/5}-1}\,e^{-4\pi i/5}\\ &=0\tag4 \end{align} $$ Solving for the positive root of $(4)$ gives $$ \begin{align} \cos(2\pi/5) &=\frac{-1+\sqrt{5}}4\\ &=\frac{\phi-1}2\tag5 \end{align} $$ Since $\cos(2\pi/5)=2\cos^2(\pi/5)-1$ and $\cos(\pi/5)$ is positive, apply $\phi^2=\phi+1$ to get $$ \cos(\pi/5)=\frac\phi2\tag6 $$ Since $\cos(x)=\sin(\pi/2-x)$, $(5)$ and $(6)$ become $$ \begin{align} \sin(\pi/10)&=\frac{\phi-1}2\tag7\\ \sin(3\pi/10)&=\frac\phi2\tag8 \end{align} $$

robjohn
  • 345,667
  • Can you access this paper : AmericanMathematical Monthly, 125(5):400–424, 2018 ? I am curious to know if you are or not mentioned in it. – Claude Leibovici Dec 11 '19 at 11:50
  • @ClaudeLeibovici: I searched for my name and did not see it. Which reference was it? – robjohn Dec 11 '19 at 12:14
  • 1
    Have a look at https://ir.lib.uwo.ca/cgi/viewcontent.cgi?article=7340&context=etd and https://www.tandfonline.com/doi/full/10.1080/00029890.2018.1420983 – Claude Leibovici Dec 11 '19 at 17:37
  • @ClaudeLeibovici: I see that $(67)$ on pg 418 is the same as this formula once we apply $e^{W(x)}=\frac{x}{W(x)}$ and $n!=\Gamma(n+1)$. My answer does predate that MAA issue, but I don't know if that means anything. – robjohn Dec 11 '19 at 18:30
  • 1
    @ClaudeLeibovici: the thesis simply quotes the MAA paper. I guess I might have been able to get a paper out of the formula. I didn't realize it wasn't common knowledge. – robjohn Dec 11 '19 at 18:38