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I have a problem understanding what the Riemann Series Theorem means. Recently I watched a magical video from Numberphile. Here they showed $\zeta(-1) = -\frac{1}{12}$. This completely boggled my mind, especially as the same is also stated in the Wikipedia article.

They did something involving adding arithmetic progressions. All I do here is based on their methods, which I think were wrong.

Let $S = 1-1+1-1+1-1+...= 0 $ or $ 1= \frac12$.
$\therefore 2S = 1$

In fact:
$\ \ \ \ \ \ S = 1-1+1-1+1-1+...=\frac12$
$\ \ \ \underline{ +S = \ \ \ \ \ \ \ 1-1+1-1+1+...=\frac12}$
$\ \ \ \ 2S =1-0+0-0+0-0+...=1$

Or:
$\ \ \ \ \ \ S = 1-1+1-1+1-1+...=\frac12$
$\ \ \ \underline{ +S=1-1+1-1+1-1+...=\frac12}$
$\ \ \ \ 2S =2-2+2-2+2-2+...=1$

The results were the same despite different methods!

What if one series starts when the other ends?

$2S=S+S$
$2S= (1-1+1-1+1-1+...)+(1-1+1-1+1-1+...)$
$2S=1-1+1-1+1-1+...$
$2S=S$
$\therefore 2=1$

Now this is completely absurd. Is it even possible to add infinite series or was all this wrong because of Riemann Series Theorem. I would appreciate if anyone would point out my mistake. Thanks in advance!

For the love of maths
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    Yes you are correct numberphile's video is absurd. What is not absurd is the theory of (analytic continuation) regularization of divergent series. We observe that the Abel regularization is translation invariant and assigns $\displaystyle\overset{(A)}\sum_{n\ge 1} n= \infty,\overset{(A)}\sum_{n\ge 1} (-1)^{n+1} = \frac12$, while the zeta regularization is not translation invariant and assigns $\displaystyle\overset{(Z)}\sum_{n\ge 1} n= \frac{-1}{12},\overset{(Z)}\sum_{n\ge 1} (-1)^{n+1} = \frac12$. The absurdity in the video and that you noticed is when mixing zeta and Abel regularization. – reuns Dec 26 '17 at 08:06
  • So $\zeta(-1)=-\frac{1}{12}$ but $1+2+3+4+5+6+7+8+...=-\frac{1}{12}$ is wrong. @reuns – For the love of maths Dec 26 '17 at 08:08
  • Yes $\sum_{n \ge 1} n = \infty$ while $\displaystyle\overset{(Z)}\sum_{n\ge 1} n= \frac{-1}{12}$ and $\displaystyle\overset{(Z)}\sum_{n\ge 1} (n-1)= \frac{-1}{12} +\frac14$, those are 3 completely different things. – reuns Dec 26 '17 at 08:10
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    The limit of the number of stupid things in this kind of videos when they are related to mathematics is $> \infty$ (just to add one more stupidity from my own). – Claude Leibovici Dec 26 '17 at 08:37

1 Answers1

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This is also known as Grandi’s series. You can check out my earlier answer on MSE.

A succinct explanation is also given on Brilliant and in Wikipedia, the latter having an explanation for the same sum which you have shown here.