What is sum: $\sum\limits_{m,n\geq1}\frac{1}{(1+mn)^2}$?

Question

What is the sum $$\sum\limits_{m,n\geq1}\frac{1}{(1+mn)^2}.$$

Answer 1

$$\begin{eqnarray*} \sum_{m,n\geq 1}\frac{1}{(1+mn)^2} = \sum_{r\geq 1}\frac{d(r)}{(r+1)^2}=\frac{1}{4}+\sum_{r\geq 2}d(r)\left(\frac{1}{r^2}-\frac{2}{r^3}+\frac{3}{r^4}-\ldots\right)\end{eqnarray*}$$ but: $$ \sum_{r \geq 2}\frac{d(r)}{r^k} = -1+\sum_{r\geq 1}\frac{(1*1)(r)}{r^k} = -1+\zeta(k)^2 $$ so:

$$\sum_{m,n\geq 1}\frac{1}{(1+mn)^2}=\frac{1}{4}+\sum_{h\geq 2}(h-1)(-1)^h\left(\zeta(h)^2-1\right).$$

Answer 2

I have an idea that might help, but I can't finish:

Note that for each positive integer $k$, the term $$\frac1{(1+k)^2}$$ occurs as many times as the number of divisors of $k$, that is $$\sum_{m,n\ge1}\frac1{(1+mn)^2}=\sum_{k=1}^\infty\frac{d(k)}{(1+k)^2}$$

where $d(k)$ is the number of divisors of $k$.