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It is famous by Ramanujan that $1+2+3+...=\frac{-1}{12}$.
But what is the value of $1+1+1=....$?

I was thinking that $2=1+1$ and $3=1+1+1$ and so on...So does this mean that $1+1+1+...=\frac{-1}{12}$.

It means that when you have infinity you can do anything and everything. Please explain this

rash
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  • The sum is famous. But is it correct? – Landuros Aug 21 '19 at 03:36
  • https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF , in particular the sentence "...where the left-hand side has to be interpreted as being the value obtained by using one of the aforementioned summation methods and not as the sum of an infinite series in its usual meaning..." So you have to understand the context here, otherwise it makes no sense – imranfat Aug 21 '19 at 03:38
  • $1+2+3+\dots \neq -\frac{1}{12}$ – Axion004 Aug 21 '19 at 03:38
  • Before you can use algebra to determine what an infinite series sums to, you need to prove that it converges. The famous sum that you’re referring to clearly does not equal $-1/12$, since it doesn’t converge. – Joe Aug 21 '19 at 03:40
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    this is obviously a different question – rash Aug 21 '19 at 06:20
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    Mine says why $1+1+1+....$ is like that because of the sum series. How is it even a duplicate of it? – rash Aug 21 '19 at 06:21

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The axioms used for this derivation don't assume that $$1 + (1+1) + (1+1+1) + (1+1+1+1) + \dots = 1 + (1 + (1 + (1 + \dots$$

In fact the freedom to arbitrarily regroup terms is not assumed for series in general, let alone tricky things like this.

J. W. Tanner
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DanielV
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