Dirichlet series
- Tolaso J Kos
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Dirichlet series
Let $\sigma(n)$ denote the sum of of all divisors of $n$, that is $\displaystyle {\rm \sigma(n)=\sum \limits_{d \mid n} d}$. Prove that for $s \in \mathbb{R} \mid s>2$ it holds that:
$$\sum_{n=1}^{\infty} \frac{\sigma(n)}{n^s} = \zeta(s) \zeta(s-1)$$
where $\zeta$ is the Riemann zeta function.
This is not a casual number thoery problem rather an analytic number theory problem. I found it very interesting and thus I am sharing it with you.
$$\sum_{n=1}^{\infty} \frac{\sigma(n)}{n^s} = \zeta(s) \zeta(s-1)$$
where $\zeta$ is the Riemann zeta function.
This is not a casual number thoery problem rather an analytic number theory problem. I found it very interesting and thus I am sharing it with you.
Imagination is much more important than knowledge.
Re: Dirichlet series
Let $\displaystyle F(z)=\sum_{m=1}^{\infty} \frac{f(m)}{m^z}$ and $\displaystyle G(z)=\sum_{n=1}^{\infty} \frac{g(n)}{n^z}$ be two complex series that converge absolutely somewhere in the complex plane then we define the convolution Dirichlet product as follows:
$$F(z)G(z) = \sum_{m=1}^{\infty} \frac{f(m)}{m^z} \sum_{n=1}^{\infty} \frac{g(n)}{n^z} = \sum_{n=1}^{\infty} \frac{h(n)}{n^z} = \sum_{n=1}^{\infty} \frac{\left ( f*g \right )(n)}{n^z} = \sum_{n=1}^{\infty} \frac{\sum \limits_{{\rm d\mid n}} f(d) g\left ( \frac{n}{d} \right )}{n^z}$$
So taking $f(n)=1, \; g(n)=n$ we have that the convolution product is actually $\sigma(n)$ thus:
$$\sum_{m=1}^{\infty} \frac{1}{m^s} \sum_{n=1}^{\infty} \frac{n}{n^{s}} = \sum_{n=1}^{\infty}\frac{\sigma(n)}{n^s} \Rightarrow \sum_{n=1}^{\infty}\frac{\sigma(n)}{n^s} = \zeta(s) \zeta(s-1)$$
that is what we wanted.
Remark: Extending the result we may obtain that for $s>a$ and $\displaystyle {\rm \sigma_a(n)=\sum \limits_{d \mid n^a} d}$ the following result holds:
$$\sum_{n=1}^{\infty} \frac{\sigma_a(n)}{n^s} = \zeta(s) \zeta(s-a)$$
$$F(z)G(z) = \sum_{m=1}^{\infty} \frac{f(m)}{m^z} \sum_{n=1}^{\infty} \frac{g(n)}{n^z} = \sum_{n=1}^{\infty} \frac{h(n)}{n^z} = \sum_{n=1}^{\infty} \frac{\left ( f*g \right )(n)}{n^z} = \sum_{n=1}^{\infty} \frac{\sum \limits_{{\rm d\mid n}} f(d) g\left ( \frac{n}{d} \right )}{n^z}$$
So taking $f(n)=1, \; g(n)=n$ we have that the convolution product is actually $\sigma(n)$ thus:
$$\sum_{m=1}^{\infty} \frac{1}{m^s} \sum_{n=1}^{\infty} \frac{n}{n^{s}} = \sum_{n=1}^{\infty}\frac{\sigma(n)}{n^s} \Rightarrow \sum_{n=1}^{\infty}\frac{\sigma(n)}{n^s} = \zeta(s) \zeta(s-1)$$
that is what we wanted.
Remark: Extending the result we may obtain that for $s>a$ and $\displaystyle {\rm \sigma_a(n)=\sum \limits_{d \mid n^a} d}$ the following result holds:
$$\sum_{n=1}^{\infty} \frac{\sigma_a(n)}{n^s} = \zeta(s) \zeta(s-a)$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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