Some formulae for Riemann Zeta

Formulae for Riemann Zeta function that allow you to see the relations between partial sums and the function value at an arbitrary complex number:

$\displaystyle \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^{s}}=\frac{1}{1^{s}}+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\dots \text { for } \Re(s)>1$

$\displaystyle \zeta(s)=\lim _{N \rightarrow \infty}\left(\sum_{n=1}^{N} \frac{1}{n^{s}}-\frac{N^{1-s}}{1-s}\right) \text { for } \Re(s)>0 \text { and } \Re(s) \neq 1$

$\displaystyle \zeta(s)=\lim _{N \rightarrow+\infty}\left(\sum_{n=1}^{N} \frac{1}{n^{s}}-\frac{N^{1-s}}{1-s}-\frac{1}{2} N^{-s}\right) \text { for } \Re(s)>-1 \text { and } \Re(s) \neq 1$

$\displaystyle \zeta(s)=\lim _{N \rightarrow+\infty}\left(\sum_{n=1}^{N} \frac{1}{n^{s}}-\frac{N^{1-s}}{1-s}-\frac{1}{2} N^{-s}+\frac{1}{12} s N^{-1-s}\right), \text {for } \Re(s)>-2$

$\displaystyle \zeta(s)=\lim _{N \rightarrow+\infty}\left(\sum_{n=1}^{N} \frac{1}{n^{s}}-\frac{N^{1-s}}{1-s}-\frac{1}{2 N^{s}}+\sum_{i=1}^{\infty} \frac{B_{2 i}}{(2 i) !} \frac{(s+2 i-2) !}{(s-1) ! N^{s+2 i-1}}\right)$

$\displaystyle \zeta(s)=\frac{1}{s-1}+\frac{1}{2}+\sum_{\ell=2}^{k} \frac{B_{\ell}}{\ell !} s(s+1) \cdots(s+\ell-2)\\ \quad\quad\quad~~~~~~~~~~~~~~~~~~~~~~~-\frac{(-1)^{k}}{k !} \int_{1}^{\infty} s(s+1) \cdots(s+k-1) x^{-s-k} B_{k}(\{x\}) d x$

$\displaystyle \text{Smoothed Sums:} \quad\quad \sum_{n=1}^{\infty} \frac{1}{n^{s}} \phi(n / N)=\zeta(s)+ N^{1-s}\int_{0}^{\infty} x^{-s} \phi(x) d x+O(1 / N)$

$\displaystyle \zeta(-s) =\frac{-B_{s+1}}{s+1} \to \quad\quad \sum_{n=1}^{\infty} n^{s} \phi(n / N)=-\frac{B_{s+1}}{s+1}+ N^{s+1}\int_{0}^{\infty} x^{s} \phi(x) d x \\\quad~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+O\left(\frac{1}{N}\right)$

Using these formulae, it’s easy to see that

$\displaystyle \zeta(-2n) =0$
$\displaystyle \zeta(-2n-1) =-\frac{B_{2n+1}}{2n+1}$
$\displaystyle \zeta(0)=-\frac{1}{2}$
$\displaystyle \zeta(-1) =-\frac{1}{12}$