What does this mean ?

[quiet]

$\lim_{\left(n \rightarrow \infty \right)} \sum_{i=1}^{i=n} i= - \dfrac{1}{12}$

I've seen it on the next page.

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

Edited October 7, 2018 by quiet

[taeto]

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

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

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

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

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

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

Edited October 7, 2018 by taeto

[quiet]

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

[taeto]

  On 10/7/2018 at 7:46 PM, quiet said:

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

Expand  

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

[quiet]

Thank you very much taeto !