<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] Let $p_1, p_2, \ldots$ be the list of primes and 
$\mathcal{E}$ be the collection of non-negative integer sequences
$( e_k )_{k\in \mathbb{Z}_{+}}$ with finitely many non-zero terms.
Evey $n \in \mathbb{Z}_{+}$ can be represented as a product of the form
$\prod\limits_{k=1}^\infty p_k^{e_k}$ for an unique $e \in \mathcal{E}$.
We have
$$\sum_{n\in\mathbb{Z}_{+}} \frac{d(n)}{n^2} 
= \sum_{e\in \mathcal{E}} 
\frac{d\left(\prod_{k=1}^\infty p_k^{e_k}\right)}{\left(\prod_{k=1}^\infty p_k^{e_k}\right)^2}
= \sum_{e\in \mathcal{E}} \prod_{k=1}^\infty \frac{e_k+1}{p_k^{2e^k}}
= \prod_{k=1}^{\infty} \left[ \sum_{e_k=0}^\infty \frac{e_k+1}{p_k^{2e^k}} \right]\\
= \prod_{k=1}^{\infty} \left[ 1 - \frac{1}{p_k^2}\right]^{-2}
= \left\{\prod_{k=1}^{\infty} \left[ 1 - \frac{1}{p_k^2}\right]^{-1}\right\}^2
= \left\{\prod_{k=1}^{\infty} \sum_{e_k=0}^{\infty} \frac{1}{p_k^{2^{e_k}}}\right\}^2\\
= \left\{\sum_{e\in \mathcal{E}} 
\frac{1}{\left(\prod_{k=1}^\infty p_k^{e_k}\right)^2}
\right\}^2
= \left\{\sum_{n\in\mathbb{Z}_{+}}\frac{1}{n^2}\right\}^2
= \zeta(2)^2
= \frac{\pi^4}{36}
$$
</s>