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This question already has an answer here:

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?

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marked as duplicate by arjafi, Pedro Tamaroff, Jasper Loy, leonbloy, Zev Chonoles Apr 7 '13 at 21:41

This question was marked as an exact duplicate of an existing question.

Duplicate of – Zev Chonoles Apr 7 '13 at 21:36
Thank you Zev! I tried to search but nothing came up. My question is answered. – Arbias Hashani Apr 7 '13 at 21:38

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.

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But the series here is not a power series at all... – Henning Makholm Apr 7 '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 7 '13 at 21:55

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