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The staff of Numberphile has shown that the sum of all the integers from $0$ to $\infty$ is $-\frac1{12}$. Recently I was looking for the sum of all the (positive) integers from $0$ to $n$ and I found that: $$\sum_{i=0}^n i=\frac{n(n+1)}{2}$$ So I decided to take the limit: $$\lim_{n\to \infty}\frac{n(n+1)}{2}$$ but that tends towards $\infty$ when I expected that to be $-\frac1{12}$!
Where did I got wrong? (the result is also confirmed by Wolfram Alpha)


marked as duplicate by vadim123, Will Jagy, Namaste, Asaf Karagila, Ivan Loh May 31 '14 at 18:39

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    $\begingroup$ You did go wrong in believing that Numberphile video, which is old history in this site of ours... $\endgroup$ – DonAntonio May 31 '14 at 18:05
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    $\begingroup$ Under the normal definition of an infinite sum, this infinite sum diverges and thus has no finite value. When people "show that the sum is $-\frac{1}{12}$", what they really mean is something along the lines of "This sum is useful in many areas of physics, and if we are to assign any meaningful finite value to it, $-\frac{1}{12}$ is the only one that makes sense. Thus we may call the 'physics sum' $-\frac{1}{12}$." But because many people are not that uptight about abuse of notation, they decide to just call it the sum anyway, even though their definition of "sum" is different. $\endgroup$ – Ivan Loh May 31 '14 at 18:38
  • $\begingroup$ Perhaps less confusion would arise if people were more careful in distinguishing what they mean by 'sum', and take care to use a different wording when they mean a different definition of 'sum', e.g. saying Cesaro sum instead of just sum when they are using Cesaro summation. $\endgroup$ – Ivan Loh May 31 '14 at 18:39
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    $\begingroup$ Honestly, I'm not into this complicated summing but if you sum a positive integer with another positive integer, it should be positive, therefore as it tends towards infinity it should value towards infinity. $\endgroup$ – Saketh Malyala Aug 8 '15 at 17:33

The numberphile video is incorrect, basically for this reason.

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    $\begingroup$ But that result is used in string theory! $\endgroup$ – Peterix May 31 '14 at 18:01

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