Last month I was calculating $\displaystyle \int_0^\infty \frac{1}{1+x^4}\, dx$ when I stumbled on the surprising identity:

$$\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right) = \frac{\pi}{\sqrt8}$$

and I knew

$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}$$

So if I could find a proof that $$\left(\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)\right)^2 = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$ then this could be a new proof that $\zeta(2)=\frac{\pi^2}{6}$. I've thought over this for almost a month and I'm no closer on showing this identity.

Note: Article on the multiplication of conditionally convergent series: http://www.jstor.org/stable/2369519

  • $\begingroup$ Thanks for the neat reference. $\endgroup$ – Andrés E. Caicedo Dec 9 '13 at 0:38
  • 1
    $\begingroup$ I believe my answer here can be adapted to answer your question. Essentially, the answer shows how to obtain $\zeta(2) = \pi^2/6$, by squaring $1-\dfrac13 + \dfrac15 \mp \cdots = \dfrac{\pi}4$. I am now too lazy to write the entire thing out here. $\endgroup$ – user17762 Dec 9 '13 at 0:41
  • $\begingroup$ Some things are so obvious that we never see them ! :-) $\endgroup$ – Lucian Dec 9 '13 at 3:24
  • 2
    $\begingroup$ Oh, and as far as that integral is concerned, $$\int_0^\infty\frac{dx}{1+x^n}=\frac{\frac\pi n}{\sin\left(\frac\pi n\right)}$$ $\endgroup$ – Lucian Dec 9 '13 at 4:16
  • 2
    $\begingroup$ Let $t=\displaystyle\frac1{1+x^n}$. It will become a Beta function, which is expressible in terms of the Gamma function. Then use the reflection formula for the latter. See my answer here. $\endgroup$ – Lucian Dec 12 '13 at 1:19

Let's have a try. $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{4n+1}=\int_{0}^{1}\frac{dx}{1+x^4},\qquad S=\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)=\int_{0}^{1}\frac{1+x^2}{1+x^4}dx,$$ $$ S = \int_{0}^{1}\frac{x+x^{-1}}{x^{-2}+x^2}\frac{dx}{x}=\int_{1}^{+\infty}\frac{z}{(2z^2-1)\sqrt{1-z^2}}\,dz = \int_{0}^{1}\frac{dt}{(2-t^2)\sqrt{1-t^2}},$$ $$ S = \int_{0}^{\pi/2}\frac{d\theta}{2-\sin^2\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta}=\frac{1}{2}\int_{\mathbb{R}}\frac{du}{2+u^2},$$ where in the last integral we used the substitution $\theta=\arctan u$. This gives: $$ S^2 = \frac{1}{8}\int_{\mathbb{R}^2}\frac{du\,dv}{(1+u^2)(1+v^2)}=\int_{0}^{1}\int_{0}^{+\infty}\frac{1}{(1+z^2)(1+x^2)}dx\,dz$$ On the other hand, $$\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\int_{0}^{1}\frac{\log y}{y^2-1}dy=\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2y^2)}dx\,dy,$$ where I learned the last equality from the Mike Spivey's note on the Luigi Pace's proof of $\zeta(2)=\frac{\pi^2}{6}$, just here. By setting $y=\frac{z}{x}$ in the last integral we get $S^2=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}$, QED. So it looks like @user17762's proof-by-squaring-the-arctangent-series and Pace's proof can be combined in order to get a very short proof of your claim.

For the sake of exposing a one-line-proof of $\zeta(2)=\frac{\pi^2}{6}$: $$\zeta(2)=\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy=\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\frac{\pi^2}{6}.$$

  • $\begingroup$ I love this. But I don't see how you used @user17762's proof. I tried following its route and I could never find an asymptotic function like the one used in it. $\endgroup$ – genepeer Dec 25 '13 at 5:22
  • 1
    $\begingroup$ The last inequality looks a lot like another one used in a recent proof (2013) by Danielle Ritelli. Thought you might like the reference. euler.genepeer.com/?p=212 $\endgroup$ – genepeer Dec 25 '13 at 5:41
  • $\begingroup$ In fact, to consider the integral $\int_{\mathbb{R}^2}\frac{dx\,dy}{(1+x^2)(1+y^2)}$ (or a slight variant) is just to consider the square of the arctangent series. @user17762 manipulates it as a double sum, here it appears as a double integral. I did not know Danielle Ritelli's proof, but our approaches are almost identical, so I think I cannot take credits for this :) $\endgroup$ – Jack D'Aurizio Dec 25 '13 at 9:12
  • $\begingroup$ I was able to prove the identity directly :) $\endgroup$ – genepeer Feb 25 '18 at 5:59

Let $a_k = (-1)^k \left(\frac{1}{4k+1} + \frac{1}{4k+3}\right)$ and $b_k = \frac{1}{(4k+1)^2} + \frac{1}{(4k+3)^2}$. The goal is to show that: $$ \left(\sum_{i=0}^\infty a_i\right)^2 = \sum_{i=0}^\infty b_i $$ The key observation that I missed on my previous attempt is that: $$ \sum_{i=0}^n a_i = \sum_{i=-n-1}^n \frac{(-1)^i}{4i+1} $$ This transformation allows me to then mimic the proof that was suggested in the comments by @user17762. \begin{align*} \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i &= \left(\sum_{i=-n-1}^n \frac{(-1)^i}{4i+1}\right)^2 - \sum_{i=-n-1}^n \frac{1}{(4i+1)^2} \\ &= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^i}{4i+1}\frac{(-1)^j}{4j+1} \\ &= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^{i+j}}{4j-4i}\left(\frac{1}{4i+1}-\frac{1}{4j+1} \right) \\ &= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^{i+j}}{2j-2i} \cdot \frac{1}{4i+1} \\ &= \frac{1}{2}\sum_{i=-n-1}^n \frac{(-1)^i}{4i+1} \sum_{\substack{j=-n-1 \\ i \neq j}}^n \frac{(-1)^j}{j-i} \\ &= \frac{1}{2}\sum_{i=-n-1}^n \frac{(-1)^i }{4i+1}c_{i,n} \\ &= \frac{1}{2}\sum_{i=0}^n a_i \,c_{i,n} \end{align*} Where the last equality follows from $c_{i,n} = c_{-i-1, n}$. Since $c_{i,n}$ is a partial alternating harmonic sum, it is bounded by its largest entry in the sum: $\left| c_{i,n} \right| \le \frac{1}{n-i+1}$. We also know that $\left|a_i\right| \le \frac{2}{4i+1}$. Apply these two inequalities to get: \begin{align*}\left| \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i \right| &\le \frac{1}{2} \sum_{i=0}^n \frac{2}{4i+1} \cdot \frac{1}{n-i+1} \\ &\le \sum_{i=0}^n \frac{1}{4n+5}\left( \frac{4}{4i+1} + \frac{1}{n-i+1} \right) \\ &\le \frac{1}{4n+5}\left( 5 + \ln(4n+1) +\ln(n+1)\right) \\ & \to 0 ~\text{ as }~ n \to \infty \end{align*} This concludes the proof. In fact, with the same idea, you can prove this general family of identities: Fix an integer $m \ge 3$, then:

\begin{align*} & \left( 1 + \frac{1}{m-1} - \frac{1}{m+1} - \frac{1}{2m-1} + \frac{1}{2m+1} + \frac{1}{3m-1} - \cdots \right)^2 \\ =& ~ \left(\sum_{i=-\infty}^\infty \frac{(-1)^i}{im+1}\right)^2 \\ =& ~ \sum_{i=-\infty}^\infty \frac{1}{(im+1)^2} \\ =& ~ 1 + \frac{1}{(m-1)^2} + \frac{1}{(m+1)^2} + \frac{1}{(2m-1)^2} + \frac{1}{(2m+1)^2} + \frac{1}{(3m-1)^2} + \cdots \\ =& ~ \left(\frac{\frac{\pi}{m}}{\sin\frac{\pi}{m}}\right)^2 \end{align*} The last equality follows from the comment by @Lucian.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.