Riemann zeta function \(\zeta(s)\) and \(\displaystyle\sum_{n=1}^\infty n=1+2+3+\cdots=-\dfrac{1}{12}\)

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:
\[\zeta(s)=\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^s}=\dfrac{1}{1^s}+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\cdots\]
for \(s\in\mathbb{C}\) such that \(\Re(s)>1\).
It can be extended to a meromorphic function with only a simple pole at \(s=1\), using analytic continuation and the following functional equation:
\[\zeta(1-s)=2^{1-s}\pi^{-s}\cos\left(\dfrac{\pi s}{2}\right)\Gamma(s)\zeta(s)\]
For \(s=2\), this gives \(\zeta(-1)=\displaystyle\sum_{n=1}^\infty n=-\dfrac{1}{2\pi^2}\zeta(2)=-\dfrac{1}{2\pi^2}\cdot\dfrac{\pi^2}{6}=-\dfrac{1}{12}\), which is a reason for assigning a finite value to the divergent sum/series (zeta function regularization). That is, \(\displaystyle\sum_{n=1}^\infty n=1+2+3+\cdots=-\dfrac{1}{12}\).
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How could adding positive integers equal a negative fraction?🤔 🔗 https://www.cantorsparadise.com/the-ramanujan-summation-1-2-3-1-12-a8cc23dea793

Ramanujan's summation: \(1+2+3+\cdots=-\dfrac{1}{12}\)

#RamanujanSummation #Ramanujan #Summation #CantorsParadise #InfiniteSum #NaturalNumbers #NegativeFraction #Fraction #NegativeNumber