3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.