Ramanujan's Natural Sums and how we get $\frac{1}{(1+1)^2}$

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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.