[math] \lim_{\left(n \rightarrow \infty \right)} \sum_{i=1}^{i=n} i= - \dfrac{1}{12} [/math]

I've seen it on the next page.

https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_⋯

Edited October 7, 20187 yr by quiet

It is a meme, see en.wikipedia.org/wiki/Meme

The fact is that [math]\sum_{n=1}^\infty n^{-s}[/math] is convergent for every complex number [math]s[/math] with [math]\mbox{Re}\ s > 1[/math].

 But the function that maps  [math]s[/math] to [math]\sum_{n=1}^\infty n^{-s}[/math] is only defined for [math]\mbox{Re}\ s > 1[/math].

So there is another function, called the Riemann zeta function  [math]\zeta[/math] which is defined for all complex numbers except  [math]1[/math], which is a pole.

The Riemann zeta function satisfies [math]\zeta(s) = \sum_{n=1}^\infty n^{-s}[/math] for all complex values [math]s[/math] with  [math]\mbox{Re}\ s > 1[/math], but not elsewhere.

The value of [math]\zeta(-1)[/math] happens to be  [math]-1/12[/math]. For [math]s=-1[/math] the expression  [math]\sum_{n=1}^\infty n^{-s}[/math] makes no sense.

Edited October 7, 20187 yr by taeto

If I say that the equation of my initial note is wrong, am I telling the truth?

14 minutes ago, quiet said:

If I say that the equation of my initial note is wrong, am I telling the truth?

Yes, the truth is that the limit does not exist.

Thank you very much taeto !