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Why does $1+2+3+\dots = {-1\over 12}$?
I found this article on Wikipedia which claims that $\sum\limits_{n=0}^\infty n=-1/12$. Can anyone give a simple and short summary on the Zeta function (never heard of it before) and why this odd result is true?
The answer is much more complicated than $\lim_{x \to 0} \frac{\sin(x)}{x}$.
The idea is that the series $\sum_{n=1}^\infty \frac{1}{n^z}$ it is convergent when $Re(z) >1$, and this works also for complex numbers.
The limit is a nice function (analytic) and can be extended in an unique way to a nice function $\zeta$. This means that
$$\zeta(z)=\sum_{n=1}^\infty \frac{1}{n^z} \,;\, Re(z) >1 \,.$$
Now, when $z=-1$, the right side is NOT convergent, still $\zeta(-1)=\frac{-1}{12}$. Since $\zeta$ is the ONLY way to extend $\sum_{n=1}^\infty \frac{1}{n^z}$ to $z=-1$, it means that in some sense
$$\sum_{n=1}^\infty \frac{1}{n^{-1}} =-\frac{1}{12}$$
and this is exactly what that means. Note that, in order for this to make sense, on the LHS we don't have convergence of series, we have a much more suttle type of convergence: we actually ask that the function $\sum_{n=1}^\infty \frac{1}{n^z}$ is differentiable as a function in $z$ and make $z \to -1$...
In some sense, the phenomena is close to the following:
$$\sum_{n=0}^\infty x^n =\frac{1}{1-x} \,;\, |x| <1 .$$
Now, the LHS is not convergent for $x=2$, but the RHS function makes sense at $x=2$. One could say that this means that in some sense $\sum_{n=0}^\infty 2^n =-1$.
Anyhow, because of the Analyticity of the Riemann zeta function, the statement about $\zeta(-1)$ is actually much more suttle and true on a more formal level than this geometric statement...