It is well known that $1+2+3+...+\infty =\frac{-1}{12} $.But if we write down $2$ as $1+1$, $3$ as $1+1+1$ and so on the sum should me written as $1+1+1+1+...+\infty$ and I don't thing has the same result as the first sum.So the thing is that that even if the sum converges to $\frac{-1}{12} $ and hence its terms can be arranged the result is clearly different, how is that possible ?(I tried to google it but nothing was useful, sorry if you think this is a dumb question)
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I don't believe so because they accept why the first summation is $-\frac{1}{12}$ Their question is about a variant on it. – Harnoor Lal Feb 24 '17 at 15:25
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Two points I have to say: one, it is not true that $$1+2+3+\dots=-\frac1{12}$$This is clearly false, and it is only meant to be true in the right context, and two, you can't manipulate terms like this. – Simply Beautiful Art Feb 24 '17 at 15:25
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Search this site for 1/12 and you will find lots of information on this infinite sum. http://math.stackexchange.com/search?q=1%2F12 – Ethan Bolker Feb 24 '17 at 15:25
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Be careful with your wording. The sum doesn't converge to $\frac{-1}{12}$ in the conventional sense, where you get the results about arrangements and whatnot. – Aweygan Feb 24 '17 at 15:25
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1"It is well known that". No. This sum is not what most people think, and definitely not a "regular" summation; and I for one does not know the actual regularization needed for this summation process to make sense and yield that value. And apparently, you may not really know it either. – Clement C. Feb 24 '17 at 15:26
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"It is well known" amongst people who watch numberphile. – Aweygan Feb 24 '17 at 15:27
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i think useful to create the tag 1/12. – cgiovanardi Feb 25 '17 at 00:40