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Alright, I thought I had seen everything but last night I saw this identity (`twas attributed to Ramanujan),

$$ 1 + 2 + 3 + 4 + \cdots = -\frac{1}{12} $$

Then I saw a proof that was seemingly correct. So alright, I believe it, hey it is no crazier than having infinities of different sizes and I finally have some closure with that fact. But then, I recalled the benchmark induction proof everyone learns,

$$ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $$

Then kicks in the remains of all those calculus courses I once took, making me thing that, hey wait! We have this,

$$ \lim_{n\rightarrow\infty} \frac{n(n+1)}{2} = \infty $$

I think in this case we said the limit does not exist or the function diverges (correct me if I am wrong!) But... but... according to the identity above,

$$ \sum_{i=1}^{n=\infty} i = -\frac{1}{12} $$

But then shouldn't,

$$ \lim_{n\rightarrow\infty} \frac{n(n+1)}{2} \stackrel{?}{=} -\frac{1}{12} $$

So what I am seeing here is that even if the limit does not converge, the sum does. Also, a long time ago I remember being told that the sum of two positive integers is always positive. Furthermore, addition is suppose to be closed under integers right? Here we not only have a negative number as a result of the sum of positive integers but a negative non-integer at that.

scribe
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    This $-1/12$ business is somewhat infamous. https://math.stackexchange.com/questions/39802/why-does-123-cdots-frac112 – AlvinL Jan 07 '21 at 07:16
  • The trouble here is that it has been a while since I have done any calculus, so I don't quite remember it rigorously but mostly the essence. Could you please answer the question in the same language I asked? Or maybe use that as a starting point. I am sorry for the trouble. – scribe Jan 07 '21 at 07:20
  • tl;dr The first equality in your post is false. $\sum _{k\in\mathbb N} a_k$ converges only if $a_k\to 0$. – AlvinL Jan 07 '21 at 07:20
  • Which equality is false? – scribe Jan 07 '21 at 07:21
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    The limit does not converge. When people write this they are referring to one of several methods for "summing divergent series" outputting the answer $- \frac{1}{12}$; the "summation" here does not have its usual meaning, and in my opinion it is quite misleading to refer to it as if it were an ordinary sum, which it isn't. – Qiaochu Yuan Jan 07 '21 at 07:23
  • If you saw the proof I think you saw, the operations they do on the infinite sums only hold if they converge. So essentially, the proof argues that if the sum did converge and could be assigned a value, it would be -\frac{1}{12}. However, the sum does not converge and cannot rigorously be assigned a value. – Stephen Donovan Jan 07 '21 at 07:26
  • @QiaochuYuan well, in that case I'd like to see an answer with more detail on that. I don't think my question is a duplicate! This isn't quite fair that it got closed. – scribe Jan 07 '21 at 07:26
  • @StephenDonovan I saw a rough sketch of the proof and I'd like to argue it but someone quoted that identity out of multiple textbooks so at that point I gave in. – scribe Jan 07 '21 at 07:28
  • @StephenDonovan If I assume the sum to converge, I can prove it to be anything I want https://www.youtube.com/watch?v=6FTwMUL69u0 –  Jan 07 '21 at 07:32
  • Could someone help me with this question being a duplicate? I am not asking why $1 + 2 + 3 + 4 + \cdots = -\frac{1}{12}$. I am saying, fine it is. Then these are the consequences, how could we reconcile them? – scribe Jan 07 '21 at 07:33
  • @Forester Well of course, anything can be derived from a false statement so naturally any such statement would be correct. – Stephen Donovan Jan 07 '21 at 07:36
  • @scribe The problem is it's a false premise: the sum doesn't converge to $-\frac{1}{12}$ so there are no consequences to reconcile. – Stephen Donovan Jan 07 '21 at 07:37
  • Then what about the textbook references and "proofs" that are given as the answer to the question it as marked a duplicate of? – scribe Jan 07 '21 at 07:41
  • @scribe True statements can only be inferred from true statements. Your supposition is false. – AlvinL Jan 07 '21 at 09:57
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    It is not my supposition, it is a supposition that is true when you define the left hand side in a funny way to make things work in particular cases. I did not know about this abuse of notation which prompted this question. I liked @QiaochuYuan's comment. – scribe Jan 07 '21 at 22:59

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