Riemann zeta function $\zeta (s)$ and ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}n=1+2+3+\cdots =-{\displaystyle \frac{1}{12}}}$

Have you ever heard that the sum of all natural numbers is $-1/12$? Of course not; this doesn't make sense in the usual sum, but using a summation method based on analytic continuation of the Riemann zeta function leads to the following result.

The Riemann zeta function is defined as:
$$\zeta (s)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^s}}={\displaystyle \frac{1}{1^s}}+{\displaystyle \frac{1}{2^s}}+{\displaystyle \frac{1}{3^s}}+\cdots }$$
for $s\in \mathbb{C}$ such that $\mathrm{\Re }(s)>1$.
It can be extended to a meromorphic function with only a simple pole at $s=1$, using analytic continuation and the following functional equation:
$$\zeta (1-s)=2^{1-s}{\pi }^{-s}\cos \left({\displaystyle \frac{\pi s}{2}}\right)\mathrm{\Gamma }(s)\zeta (s)$$
For $s=2$, this gives $\zeta (-1)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}n=-{\displaystyle \frac{1}{2{\pi }^2}}\zeta (2)=-{\displaystyle \frac{1}{2{\pi }^2}}\cdot {\displaystyle \frac{{\pi }^2}{6}}=-{\displaystyle \frac{1}{12}}}$, which is a reason for assigning a finite value to the divergent sum/series (zeta function regularization). That is, ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}n=1+2+3+\cdots =-{\displaystyle \frac{1}{12}}}$.
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