# $\sum_k (-1)^k \frac{\tau(2k+1)}{2k+1}$

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?

• $\pi$ never ceases to amaze with the identities it appears in. – MSDG Sep 2 '18 at 17:27
• I'm amazed that that is convergent. – Angina Seng Sep 2 '18 at 17:36
• @JackD'Aurizio How do you use Dirichlet's test here? $\tau(2k+1)/(2k+1)$ is not decreasing. – Robert Israel Sep 2 '18 at 17:56
• @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 $\frac{1}{k^{1-\varepsilon}}$. – Jack D'Aurizio Sep 2 '18 at 18:36
• Not surprising maybe, but replace $\tau(2k+1)$ by $\tau(2k+1) + (-1)^k$ and it doesn't converge. – 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 $$\sum_{n\geq 0} \frac{\tau(2n+1)(-1)^n}{2n+1} = \sum_{m\geq 0}\frac{(-1)^m}{2m+1}\cdot\frac{\pi}{4} = \left(\frac{\pi}{4}\right)^2.$$