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?

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    $\begingroup$ 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. $\endgroup$ – Julian Rosen Oct 2 '17 at 15:37
  • $\begingroup$ I don't think you'll find anything simpler than starting from the partial fraction decomposition of $\frac{1}{e^{2i \pi z}-1}$. $\endgroup$ – reuns Oct 2 '17 at 15:44
  • $\begingroup$ The LHS is integral of 1/(1+x^2) which is Arc tan x for x=1. RHS is Fourrier expansion of f(x) =x^2 for x=Pi. may be finding a relation between the integrants help. $\endgroup$ – sirous Oct 2 '17 at 17:22
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    $\begingroup$ 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 :-/ $\endgroup$ – tired Oct 4 '17 at 23:48
  • $\begingroup$ to give some more detail: the above follows from the fact that you statement is equivalent to $$(\sum_{k=1,..,4}\chi_1(k)\zeta(1,k/4))^2=\tfrac{3}{4}\sum_{k=1,..,4}\chi_1(k)\zeta(2,k/4) $$ maybe some number thoerist can bring it home from here (the lhs involves some limits). Afterwards we apply some polygamma identities and are "done" $\endgroup$ – tired Oct 5 '17 at 17:48

Here is a proof of the identity $$\left(\sum_{n \geq 0} \frac{(-1)^n}{(2n+1)}\right)^2 = \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}$$ without knowing the value of either series.

Consider the double integral \begin{align*} I &= \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx. \end{align*} On one hand, we observe \begin{align*} \frac{1}{(1+x^2)(1+y^2)} &= \sum_{n \geq 0} \left(-x^2 \right)^n \sum_{m \geq 0} \left(-y^2 \right)^m, \quad |x|,|y| < 1. \end{align*} Then, we get that \begin{align*} I &= \int_{0}^1 \int_{0}^1 \sum_{n \geq 0} \left(-x^2 \right)^n \sum_{m \geq 0} \left(-y^2 \right)^m \ dy \ dx \\ &=\sum_{n \geq 0}\sum_{m \geq 0} \int_{0}^1 \int_{0}^1 \left(-x^2 \right)^n \left(-y^2 \right)^m \ dy \ dx \\ &=\sum_{n \geq 0}\sum_{m \geq 0} \frac{(-1)^n}{2n+1} \frac{(-1)^m}{2m+1} \\ &=\left(\sum_{n \geq 0} \frac{(-1)^n}{2n+1} \right)^2. \end{align*} The interchanging of sum and integral done here can be justified by the Dominated Convergence Theorem.

Now, we make the substitution $x \mapsto 1/x,$ and $y \mapsto 1/y$ to $I,$ resulting in \begin{align*} I &= \int_{1}^{\infty}\int_{1}^{\infty} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx, \end{align*} This implies \begin{align*} I &= \frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx. \end{align*} Make another substitution $y=tx$ with $dy = x \ dt$ to see \begin{align*} I &= \frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{x}{(1+x^2)(1+t^2x^2)} \ dt \ dx \\ &=\frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{x}{(1+x^2)(1+t^2x^2)} \ dx \ dt, \end{align*} in which the interchanging of the integral is justified by Tonelli's Theorem. Now, using the partial fractions identity, $$\frac{x}{(1+x^2)(1+t^2x^2)} = \frac{2 x}{\left(2-2 t^2\right) \left(x^2+1\right)}-\frac{2 t^2 x}{\left(2-2 t^2\right) \left(t^2 x^2+1\right)},$$ and integrating with respect to $x,$ we get \begin{align*} I &= \frac{1}{4}\int_{0}^{\infty} \frac{\log(t)}{t^2-1} \ dt. \end{align*} Now, observe \begin{align*} \int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt + \int_{1}^{\infty} \frac{\log(t)}{t^2-1} \ dt= \int_{0}^{\infty} \frac{\log(t)}{t^2-1} \ dt. \end{align*} We see that \begin{align*} \int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt &= \int_{1}^{\infty} \frac{\log(t)}{t^2-1} \ dt \end{align*} by making a change of variables $t \mapsto 1/t$ on the second integral. Thus, \begin{align*} I &= \frac{1}{2} \int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt. \end{align*} Now, convert the integrand into a geometric series \begin{align*} \frac{\log(t)}{t^2-1} &= - \sum_{n \geq 0} \log(t) t^{2n}, \quad t \in (0,1), \end{align*} and we see that \begin{align*} I &= -\frac{1}{2} \int_{0}^{1} \sum_{n \geq 0} \log(t) t^{2n} \ dt \\ &= -\frac{1}{2} \sum_{n \geq 0} \int_{0}^{1} \log(t) t^{2n} \ dt \\ &= \frac{1}{2} \sum_{n \geq 0} \frac{1}{(2n+1)^2}, \end{align*} in which the interchanging of sum and integral is justified by Beppo Levi's Monotone Convergence Theorem. Observing that \begin{align*} \sum_{n \in \mathbb{N}} \frac{1}{n^2} &= \sum_{n \in \mathbb{N}} \frac{1}{(2n)^2}+\sum_{n \geq 0} \frac{1}{(2n+1)^2} \\ &= \frac{1}{4}\sum_{n \in \mathbb{N}} \frac{1}{n^2}+\sum_{n \geq 0} \frac{1}{(2n+1)^2}, \end{align*} we find \begin{align*} I &= \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}. \end{align*} Hence, we have $$\left(\sum_{n \geq 0} \frac{(-1)^n}{(2n+1)}\right)^2 = \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}.$$

Remark For those that may be interested, these types of integral calculations to evaluate series may be found in my joint paper with Daniele Ritelli. See https://www.ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3/

  • 1
    $\begingroup$ Nice / clever use of integrals. +1 $\endgroup$ – Paramanand Singh Sep 19 '19 at 23:00

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.

  • $\begingroup$ @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 $x$. $\endgroup$ – Paramanand Singh Oct 3 '17 at 15:22
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos Oct 3 '17 at 15:25
  • $\begingroup$ @JoséCarlosSantos i posted my approach as answer. if it doesn't fit your requirements, let me know and i delete it timmediatly $\endgroup$ – tired Oct 7 '17 at 14:20

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


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

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

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


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.

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

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