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Consider the zeta function $\zeta(s)= \sum \limits_{n=1}^{\infty} \frac{1}{n^s}$.

It is established that $ \zeta(-1) = -\frac{1}{12}$.

Reference (Equation 90)

Then we have $ \zeta(-1) = \sum \limits_{n=1}^{\infty} \frac{1}{n^{-1}}= 1+2+3+4 + ... = -\frac{1}{12}$.

But of course this series is divergent. So what is the problem here?

AXH
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1 Answers1

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Power Series are only valid inside a radius of convergence. For example, the geometric series

$$1 + z^2 + z^4 + z^6 + \cdots \equiv \frac{1}{1-z^2}$$

for all $|z| < 1$. The identity does not hold for $|z| > 1$, and the identity may or may not hold when $|z|=1$. You need to read about analytic continuation.

Fly by Night
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    But the series here is not a power series at all... – hmakholm left over Monica Apr 07 '13 at 21:44
  • @HenningMakholm It can be realised as the evaluation of a power series outside of its radius of convergence. Just like we have

    $$1+z+z^2+z^3+\cdots \equiv \frac{1}{1-z}$$

    for all $|z|<1$, and then we evaluate at $z=2$ to get the "answer"

    $$1 + 2 + 4 + 8 + \cdots = -\frac{1}{2}$$

     – Fly by Night Apr 07 '13 at 21:55
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    @FlybyNight that's great for the power series $1+z+z^2+...$ but $\sum \frac{1}{n^s}$ is not a power series. It doesn't have a radius of convergence in the usual sense. – ziggurism Nov 23 '19 at 15:26
  • @FlybyNight; outside the radius of convergence doesn't mean anything, but for non-converging sums to infinity with finite answers, yours is $1+2^\infty$. – JMP Aug 17 '20 at 07:06