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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?

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marked as duplicate by Pedro Tamaroff, Ross Millikan, Eric Naslund Jun 11 '12 at 17:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Do you know what analytic continuation is? –  Pedro Tamaroff Jun 11 '12 at 17:16
    
@PeterTamaroff: If this is something like the computation of $\lim\limits_{x\to 0}\sin(x)/x=1$ then yes; otherwise I'm afraid not. –  Christian Ivicevic Jun 11 '12 at 17:20
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Not really. writing $$\sum_{n=1}^\infty n = -\frac{1}{12}$$ is a cool way of saying that $\zeta(-1)=-\frac{1}{12}$. $$\zeta \left( s \right) = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^s}}}} $$ makes sense only for $\Re(s)>1$. When we look at $\Re(s) < 1$ we need to "redefine" it. That's where analytic continuation appears. –  Pedro Tamaroff Jun 11 '12 at 17:25
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Dear Christian, This question has already been asked and answered: math.stackexchange.com/q/39802/221 Do these answers tell you what you want to know? Regards, –  Matt E Jun 11 '12 at 17:30
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@MattE: During creation of my thread the search didn't reveal any duplicate topic, but thanks for the ref to your answer. I'll be fine with it, although it takes its time to be understood. Could you link your other answer as an answer here? (Or shall we close this topic?) And thanks to you, Peter, too. –  Christian Ivicevic Jun 11 '12 at 17:33
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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...

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