TITLE: $\sum_k (-1)^k \frac{\tau(2k+1)}{2k+1}$
QUESTION [10 upvotes]: 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?

REPLY [8 votes]: 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
$$ \sum_{n\geq 0} \frac{\tau(2n+1)(-1)^n}{2n+1} = \sum_{m\geq 0}\frac{(-1)^m}{2m+1}\cdot\frac{\pi}{4} = \left(\frac{\pi}{4}\right)^2.$$
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