# Sum of all the positive integers problem [duplicate]

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 LohMay 31 '14 at 18:39

• 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. – Ivan Loh May 31 '14 at 18:38