Prove that $$\sum_{n=1}^\infty d(n)^2n^{-s}=\zeta(s)^4/\zeta(2s)$$ for $\sigma>1$

what i did:

I already proved this formally, that is, without considering convergence. I use euler products, that is, theorem 1.9 in Montgomerys multiplicative number theory:

If f is multiplicative, and $$\sum \vert f(n)\vert n^{-\sigma}<\infty$$ Then $$\sum_{n=1}^\infty f(n)n^{-s}=\prod_{p\in\mathbf{P}}(\sum_{n=0}^{\infty}f(p^n)p^{-ns}).$$ I first prove that $d$ is a multiplicative function. Then i apply Euler-products and after some technicalities, the result pops up.

However, my problem is the assumption for the Euler product. My naive bound for the divisor function is $d(n)<2\sqrt{n}$, with the rough argument that $d>\sqrt{n}$ is a divisor if and only if $n/d$ is a divisor $d<\sqrt{n}$. But this is not good enough , since this bound only lets me apply the Euler product form for $\sigma>2$.

I found some rather complicated bounds for the divisor functions on the internet, but since this is an early exercise in the montgomery Multiplicative number theory book (1.3.1 exercise 5) i doubt thats what i should use.

The post beneath consider the same problem, but it solved what i already solved, and ignore the convergence part:

Dirichlet series generating function


I found a solution myself.

We use Montgomery theorem 1.8 :

If $$\alpha(s)=\sum_{n\in\mathbb{N}}f(n) n^{-s}\quad \beta(s)=\sum_{n\in\mathbb{N}}g(n) n^{-s}$$ converges absolutely, then their product converges to $$\gamma(s)=\sum_{n\in\mathbb{N}}h(n) n^{-s}$$ Where $h(n)$ is the Dirichlet product $f*g\; (n)$

Indeed, $$\alpha(s)=\sum_{n\in\mathbb{N}}d(n)n^{-s}$$

converges absolutely for $\sigma>1$, so $$\alpha(s)^2=\sum_{n\in\mathbb{N}}d*d(n)n^{-s}$$ converges as well, due to the theorem above.

But since $d(D)d(n/D)\geq d(n)$ we get $$d*d(n)=\sum_{D\vert n}d(D)d(n/ D)\geq \sum_{D\vert n}d(n)=d(n)^2$$ and hence the result follows by direct comparison.


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