# Proving $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)$

The equality$$\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)\tag{1}$$follows from the fact that the sum of the first series is $\dfrac\pi4$, whereas the sum of the second one is $\dfrac{\pi^2}6$.

My question is: can someone provide a proof that $(1)$ holds without using this?

• I think the $\chi$- function can help. – Michael Rozenberg Oct 2 '17 at 10:33
• Interesting! I like questions like this :) (+1) – hypergeometric Oct 2 '17 at 15:29
• The equality can be rewritten $\int_0^1\int_0^1\frac{dx\,dy}{(1+x^2)(1+y^2)}=\frac{3}{8}\int_0^1\int_0^1\frac{ds\,dt}{1-st}$. Conjecturally, an equality like this should admit a proof using only additivity, change of variables, and Stokes' theorem, though I'm not sure how one would find such a proof. – Julian Rosen Oct 2 '17 at 15:37
• I don't think you'll find anything simpler than starting from the partial fraction decomposition of $\frac{1}{e^{2i \pi z}-1}$. – reuns Oct 2 '17 at 15:44
• i can somehow wrestle it down to the fact that $\cot(\pi x)^2=1/(2 \sin(\pi x))^2$ for $x=1/4$ but it somehow feels like cheating because this fact will also give me both sides seperatly :-/ – tired Oct 4 '17 at 23:48

If we put $$f(t) =\sum_{n=1}^{\infty}\frac{\sin nt} {n}$$ then $$f^{2}(t)=\sum_{n=1}^{\infty}\frac{\sin^{2}nt}{n^{2}} +\text{(terms containing }\sin nt\sin mt)$$ and integrating this term by term with respect to $t$ over $[-\pi, \pi]$ should give us $$\pi\sum_{n=1}^{\infty}\frac{1}{n^{2}}$$ and therefore we see that the RHS of the equation in question is $$\frac{3}{8\pi}\int_{-\pi}^{\pi}f^{2}(t)\,dt$$ This needs to be proved to be equal to $f^{2}(\pi/2)$. The function $f(t)$ is given in closed form as $$f(t) = \begin{cases} \dfrac{\pi - t} {2}, 0<t\leq \pi\\ 0,t=0\\ -\dfrac{\pi +t} {2}, - \pi\leq t<0 \end{cases}$$ and $f(t+2\pi)=f(t)$. So this works out fine.

• @JoséCarlosSantos: thanks for the appreciation. The above solution does use the fact that the LHS is $\pi^{2}/16$. And I am not sure if there is any direct approach to square the series for $\arctan x$ and relate it to series $\sum y^{n} /n^{2}$ for some function $y$ of $x$. – Paramanand Singh Oct 3 '17 at 15:22
• If your solution had not used the fact that the LHS is equal to $\frac{\pi^2}{16}$, then I would have marked it as the accepted answer. And I will, if nothing better comes along. – José Carlos Santos Oct 3 '17 at 15:25
• @JoséCarlosSantos i posted my approach as answer. if it doesn't fit your requirements, let me know and i delete it timmediatly – tired Oct 7 '17 at 14:20

I found a proof relying on some results from number theory, which hopefully fits your requirements

Claim:

$$\left(\sum_{n \geq 1}\frac{(-1)^n}{2n+1}\right)^2=\frac{3}{8}\sum_{n\geq1}\frac{1}{n^2}$$

We define

• $\chi_l(n)$ as the l'th Dirichlet character $\text{mod}\,4$
• $L(\chi,s)\equiv\sum_{n\geq1}\tfrac{\chi(n)}{n^s}$ is a Dirichlet-$L$ sum
• $\zeta(s,q)\equiv\sum_{n\geq0}\tfrac{1}{(n+q)^s}$ is a Hurwitz zeta function ($q=1$ gives Riemanns $\zeta(s)$)

Now we recoginze that the orignal problem can be reformulated as follows

$$L^2(1,\chi_1)=\frac{1}{2}L(2,\chi_0) \quad(*)$$

Proof:

By virtue of the identity (this holds for general characters $\text{mod}\,\,a$)$L(\chi,s)=\sum_{b\leq a}\chi(b)\zeta(s,\frac ba)$ and the explict values of the character table we get for the left hand side of $(*)$ $$\frac{1}{16}\left(\zeta(1,\tfrac{1}{4})-\zeta(1,\tfrac{3}{4})\right)^2$$

where the limit $s\rightarrow 1$ is implicitly taken. By the Stieltjes expansion of the Hurwitz Zeta function this equal to

$$\frac{1}{16}(\psi_0(1/4)-\psi_0(3/4))^2=\frac{\pi^2}{16}\cot^2(\frac{\pi}{4}) \quad(**)$$

where the equality is a consequence of the reflection formula for the Polygamma function $\psi_n(z)$.

On the other hand, for the rhs of $(*)$ we can write by the nearly the same token (here no limiting procedure is necessary)

$$\frac{1}{2\cdot16}\left(\psi_1(1/4)+\psi_1(3/4)\right)=\frac{\pi^2}{2\cdot16}\frac{1}{\sin^2(\frac{\pi}4)}\quad(***)$$

Now since $\cos^2(\frac{\pi}{4})=\frac{1}{2}$, $(**)=(***)$ and therefore $(*)$ is proven

Since i'm a theoretical physicist my knowledge of number theory is close to zero. I guess a more experienced person could conclude here much faster using some general theorems of $L$-functions or Dirichlet convolution.

• Your knowledge of number theory does not appear to be close to zero (as far as this answer indicates). Nice approach. +1 – Paramanand Singh Oct 7 '17 at 15:14
• @ParamanandSingh thanks... i think the correct formulation would be: all i know about nt is contained in this post ;-) – tired Oct 7 '17 at 15:22