I found this article on Wikipedia which claims that ∞∑n=0n=−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?
Answer
The answer is much more complicated than limx→0sin(x)x.
The idea is that the series ∑∞n=11nz 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 ζ. This means that
ζ(z)=∞∑n=11nz;Re(z)>1.
Now, when z=−1, the right side is NOT convergent, still ζ(−1)=−112. Since ζ is the ONLY way to extend ∑∞n=11nz to z=−1, it means that in some sense
∞∑n=11n−1=−112
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 ∑∞n=11nz is differentiable as a function in z and make z→−1...
In some sense, the phenomena is close to the following:
∞∑n=0xn=11−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 ∑∞n=02n=−1.
Anyhow, because of the Analyticity of the Riemann zeta function, the statement about ζ(−1) is actually much more suttle and true on a more formal level than this geometric statement...
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