I was watching the Numberphile video regarding Ramanujan's proof for $1+2+3+\cdots = \frac{-1}{12},$ and am a bit confused. Please note that I have looked online for a few hours (including Math Stack Exchange), and could not find the answer I was looking for. So, my apologies if someone else has already asked this question.

However, at one point in the video, the speaker claims that for the Geometric Series, which has the form

$$ \sum ar^n = \frac{a}{1-r}, \ \ \text{for}\ |r|<1,$$

can have $r = -1$ and take the form

$$ 1 - 1 + 1 - 1 \cdots = 1 - 2 + 3 - 4 \cdots,$$

which is "exactly what Ramanujan wanted".

I'm fine with the idea except for one thing. To my knowledge, the Geometric Series converges to the limit $\frac{a}{1-r}$ under the condition that $|r| < 1.$ In the case of $r = -1, |r|\nless 1,$ and thus this should not work. To me, it doesn't seem like we should be able to 'just put a $-1$ there because it looks right', when in fact it doesn't converge. Now, this could be a really silly question with a really simple answer regarding some series identities that I am not aware of. Anyhow, I would really appreciate some light on the subject.

  • $\begingroup$ You are right that the geometric series converges only for $\lvert r \rvert < 1.$ the series with $r=-1$ does not converge. None of the summation methods for either $1-1+1-1+\dotsb$ nor $1+2+3+\dotsb$ will show them equal; they are not. $\endgroup$ – ziggurism Dec 26 '17 at 3:26
  • $\begingroup$ You are right: it is of course wrong to plug in $r=1$ when the very theorem being used says you can't do this. $\endgroup$ – Randall Dec 26 '17 at 3:27
  • $\begingroup$ Really confused how you go from $1-1+1-1+\dots$ to $1-2+3-4+\dots$ $\endgroup$ – Simply Beautiful Art Dec 26 '17 at 3:30
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    $\begingroup$ If you want some justification for$$\sum_{n=0}^\infty(-1)^n\stackrel?=\frac1{1-(-1)}=\frac12$$you could pick at some ideas, such as analytic continuation, Abel summation, Cesaro summation, etc. For example, Abel summation here means that:$$\sum_{n=0}^\infty(-1)^n=\lim_{r\downarrow-1}\sum_{n=0}^\infty r^n=\lim_{r\downarrow-1}\frac1{1-r}=\frac12$$ $\endgroup$ – Simply Beautiful Art Dec 26 '17 at 3:33
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    $\begingroup$ This can't be understood in terms of convergence of series, it's a summation method for divergent series. See Ramanujan summation $\endgroup$ – Robert Israel Dec 26 '17 at 3:34

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