General Form for $\displaystyle \sum_{n=1}^{\infty}\frac{d\left ( kn \right )}{n^2}$

Question

The function d(x) gives the number of divisors of x. "k" is a positive integer. In Mathematica, I think, d(x) is implemented as DivisorSigma[0,x]. If you know of such a General Form or can point me to a mathematical paper that discusses such infinite series, please pass it along. I have investigated k=2048 and found the sum approximately 25.0286 (k a power of 2 in this case) and k=2017 with an approximate sum of 5.41158 (k a prime number this time). Both approximations were a result of finding the partial sum of a million terms using SAGE. David H. Bailey has published much about similar infinite sums - but I can not find this one anywhere. It would not surprise me if the final answer has $\pi$ in it - like so many infinite series I've played with lately.

Answer 1

Working with the special case presented by the OP we take $q\ge 1$ and seek to evaluate

$$L(s) = \sum_{n\ge 1} \frac{\tau(2^q n)}{n^s}.$$

Introduce $$Q(s) = \sum_{m\ge 0} \frac{\tau(2m+1)}{(2m+1)^s}.$$

This is $$Q(s) = \zeta(s)^2 - \sum_{m\ge 1} \frac{\tau(2m)}{(2m)^s} = \zeta(s)^2 - \sum_{k\ge 1} \sum_{m\ge 0} \frac{\tau(2^k(2m+1))}{(2^k(2m+1))^s} \\ = \zeta(s)^2 - \sum_{k\ge 1} \frac{k+1}{2^{ks}} \sum_{m\ge 0} \frac{\tau(2m+1)}{(2m+1)^s}$$

which yields

$$Q(s) \frac{1}{(1-1/2^s)^2} = \zeta(s)^2 \quad\text{or}\quad Q(s) = (1-1/2^s)^2 \zeta(s)^2.$$

We thus obtain for $L(s)$

$$L(s) = Q(s) \sum_{k\ge 0} (q+k+1) \frac{1}{2^{ks}} \\ = Q(s) \left(\frac{q}{(1-1/2^s)} + \frac{1}{(1-1/2^s)^2}\right).$$

This finally yields

$$L(s) = \zeta(s)^2 (q(1-1/2^s) + 1).$$

In particular we have for the example by the OP the value

$$\zeta(2)^2 (11\times 3/4 + 1) = \frac{\pi^4}{36} \frac{37}{4} = \frac{37\pi^4}{144} \approx 25.02872479.$$

Remark. Treating the general case

$$L_p(s) = \sum_{n\ge 1} \frac{\tau(p^q n)}{n^s}$$

with $p\ge 2$ a prime we obtain by the same calculation that $$Q_p(s) = (1-1/p^s)^2 \zeta(s)^2 $$

and hence

$$L_p(s) = \zeta(s)^2 (q(1-1/p^s) + 1).$$

We get for $q=1$ and $p=2017$

$$\zeta(2)^2 ((1-1/2017^2) + 1) = \frac{\pi^4}{36}\frac{2\times 2017^2-1}{2017^2} \approx 5.411615506.$$