Express the Riemann zeta function as the series with the divisor function $d(n)$

Question

How do I show this? $$\zeta(s)^2 = \sum_{n=1}^\infty \frac{d(n)}{n^s}$$

Here, $\zeta(s)$ is the Riemann zeta function, and $d(n)$ is the number of positive divisors of $n$.

This looks like the kind of thing that would be an open problem or something but it's an exercise in a textbook.

Answer 1

you can write: $$ \frac{d(n)}{n^s}=\sum_{pq=n}\frac{1}{(pq)^s} $$

and try to invert the sums to obtain: $$\sum_{p=1}^{+\infty}\sum_{q=1}^{+\infty} \frac{1}{(pq)^s}$$

Answer 2

This result has little to do with the Riemann zeta function per ce. It is merely a consequence of the following result:

$$\sum_{n=1}^\infty\sum_{m=1}^\infty a_{mn} = \sum_{k=1}^\infty d(k)a_k$$

under the assumption that all the sums converge absolutely.

The proof is fairly simple, the sum on the left hand side can be written $\sum_{k=1}^\infty c_k a_k$ where $c_k$ counts how many times $a_k$ is repeated in sum. This is nothing but the number of solutions to $mn = k$ which is the number of divisors of $k$.

Using $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \implies \zeta^2(s) = \sum_{n=1}^\infty\sum_{m=1}^\infty \frac{1}{(nm)^s}$ we can apply the theorem above with $a_k = \frac{1}{k^s}$ to get the desired result.