The identity $$ \sum_{k=0}^\infty (-1)^k \frac{\tau(2k+1)}{2k+1} = \frac{\pi^2}{16}$$ (where $\tau$ is the number-of-divisors function) has come up in OEIS sequence A222068. Surely this is "well-known"? Can anybody supply a reference?

  • $\begingroup$ $\pi$ never ceases to amaze with the identities it appears in. $\endgroup$ – MSDG Sep 2 '18 at 17:27
  • $\begingroup$ I'm amazed that that is convergent. $\endgroup$ – Angina Seng Sep 2 '18 at 17:36
  • 1
    $\begingroup$ @JackD'Aurizio How do you use Dirichlet's test here? $\tau(2k+1)/(2k+1)$ is not decreasing. $\endgroup$ – Robert Israel Sep 2 '18 at 17:56
  • $\begingroup$ @RobertIsrael: all right, not a direct application of Dirichlet's test, but on the other hand it is not surprising that $\sum_{k\geq 0}(-1)^k b_k$ is convergent if $b_k$ (at least on average) behaves like $\frac{1}{k^{1-\varepsilon}}$. $\endgroup$ – Jack D'Aurizio Sep 2 '18 at 18:36
  • $\begingroup$ Not surprising maybe, but replace $\tau(2k+1)$ by $\tau(2k+1) + (-1)^k$ and it doesn't converge. $\endgroup$ – Robert Israel Sep 2 '18 at 18:42

The LHS is the square of the Dirichlet series $$ L(\chi_4,s)=\sum_{n\geq 1}\frac{\chi_4(n)}{n^s} $$ evaluated at $s=1$, where $\chi_4$ is the non-principal Dirichlet character $\!\!\pmod{4}$.
On the other hand $$ L(\chi_4,1) = \sum_{n\geq 0}\frac{(-1)^n}{2n+1} = \int_{0}^{1}\frac{dx}{1+x^2} = \frac{\pi}{4}.$$ This argument relies on recognizing $(-1)^n \tau(2n+1)$ as $\chi_4*\chi_4$. As an alternative, by applying straightforward manipulations to $$ \sum_{n\geq 1} \tau(n) x^n =\sum_{m\geq 1}\frac{x^m}{1-x^m} $$ we reach $$ \sum_{n\geq 0} \tau(2n+1) x^{2n}(-1)^n = \sum_{m\geq 0}\frac{(-1)^m x^{2m}}{1+x^{4m+2}} $$ and by integrating both sides on $(0,1)$ we get $$ \sum_{n\geq 0} \frac{\tau(2n+1)(-1)^n}{2n+1} = \sum_{m\geq 0}\frac{(-1)^m}{2m+1}\cdot\frac{\pi}{4} = \left(\frac{\pi}{4}\right)^2.$$

  • $\begingroup$ I'm accepting this one, but I'd still like to see a reference if that exists. $\endgroup$ – Robert Israel Sep 6 '18 at 6:44

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.