<s>[INST] How find this sum $\frac{1}{1^2}+\frac{2}{2^2}+\frac{2}{3^2}+\frac{3}{4^2}+\frac{2}{5^2}+\frac{4}{6^2}+\cdots+\frac{d(n)}{n^2}+\cdots$ Question:

Find the value
  $$\dfrac{1}{1^2}+\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{3}{4^2}+\dfrac{2}{5^2}+\dfrac{4}{6^2}+\cdots+\dfrac{d(n)}{n^2}+\cdots$$

where $d(n)$ is The total number of positive divisors of $n$
I think we can use 
$$\zeta{(2)}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}+\cdots=\dfrac{\pi^2}{6}$$
I know If the prime factorization of  is given by
$$n=p^{a_{1}}_{1}\cdot p^{a_{2}}_{2}\cdots p^{a_{n}}_{n}$$
then the number of positive divisors of  is
$$d(n)=(a_{1}+1)(a_{2}+1)\cdots(a_{n}+1)$$
But follow is very ugly,I don't understand @Nate idea(my English is poor),can you post detail? Thank you 
 [/INST] Since $d(n) = \sum_{d|n}1$ you can write
$$\sum_{n=1}^{+\infty} \frac{d(n)}{n^2} = \sum_{n=1}^{+\infty} \sum_{d|n} \frac{1}{n^2}$$
Now, the set $\{ (d,n) \in \mathbb{N}^2 : d|n \}$ can be identified with the set $\{ (k, d)\in \mathbb{N}^2  \}$ by the map $(k,d) \mapsto (d, kd)$ (so $n = kd$). This means that
$$\sum_{n=1}^{+\infty} \sum_{d|n} \frac{1}{n^2}=
\sum_{d=1}^{+\infty} \sum_{k=1}^{+\infty} \frac{1}{(kd)^2} = 
\sum_{d=1}^{+\infty} \frac{1}{d^2} \sum_{k=1}^{+\infty} \frac{1}{k^2} = \left(\frac{\pi^2}{6} \right)^2$$
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