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

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

Answer 1

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

$$\sum_{n=1}^{+\infty} \sum_{d|n} \frac{1}{n^2}= \sum_{d=1}^{+\infty} \sum_{k=1}^{+\infty} \frac{1}{(kd)^2} = \sum_{d=1}^{+\infty} \frac{1}{d^2} \sum_{k=1}^{+\infty} \frac{1}{k^2} = \left(\frac{\pi^2}{6} \right)^2$$

Answer 2

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 $$\sum_{n\in\mathbb{Z}_{+}} \frac{d(n)}{n^2} = \sum_{e\in \mathcal{E}} \frac{d\left(\prod_{k=1}^\infty p_k^{e_k}\right)}{\left(\prod_{k=1}^\infty p_k^{e_k}\right)^2} = \sum_{e\in \mathcal{E}} \prod_{k=1}^\infty \frac{e_k+1}{p_k^{2e^k}} = \prod_{k=1}^{\infty} \left[ \sum_{e_k=0}^\infty \frac{e_k+1}{p_k^{2e^k}} \right]\\ = \prod_{k=1}^{\infty} \left[ 1 - \frac{1}{p_k^2}\right]^{-2} = \left\{\prod_{k=1}^{\infty} \left[ 1 - \frac{1}{p_k^2}\right]^{-1}\right\}^2 = \left\{\prod_{k=1}^{\infty} \sum_{e_k=0}^{\infty} \frac{1}{p_k^{2^{e_k}}}\right\}^2\\ = \left\{\sum_{e\in \mathcal{E}} \frac{1}{\left(\prod_{k=1}^\infty p_k^{e_k}\right)^2} \right\}^2 = \left\{\sum_{n\in\mathbb{Z}_{+}}\frac{1}{n^2}\right\}^2 = \zeta(2)^2 = \frac{\pi^4}{36} $$