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I have read online, that one can show that $\sum_{n=1}^{\infty}n = -\frac{1}{12}$.

But isn't this a Riemann Series of the form $\sum_{n=1}^{\infty} \frac{1}{n^p}$, where $p=-1$.

And if so, can't you prove that

$ \lim_{N \to \infty} \int^N \frac{1}{x^{-1}} = \lim_{N \to \infty} \frac{x^2}{2} = \infty$,

thus, by the integral test for convergence, the series diverges? Where is my mistake?

user84413
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Aaron Wild
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  • ...if this is a serious question, you must look into analytic continuation. Either that, or string theory. You cannot use calc 2 to prove it. There is a tricky way to prove with sums which can be found here – Derek Orr Aug 14 '14 at 23:27
  • @IanMateus My question isn't why $1+2+3+... = -\frac{1}{12}$, but why the integral test gives a 'WRONG' result in this case. I am realy not interested in a prove, but why the prove that it the statement is wrong itself seems to be 'wrong'. – Aaron Wild Aug 14 '14 at 23:28
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    Short answer: you are right, this does not converge and the value $-1/12$ is instead associated to $\zeta(-1)$, a "extended version" of $p$-series. This abuse of notation was exhaustively addressed on his site before. As dylan7 noted, the integral test cannot be applied to this test blindly (as this series is not monotone decreasing) but you can easily show by area comparison that the sum is divergent. – Ian Mateus Aug 14 '14 at 23:29
  • You are correct that the series diverges, as Ian points out, and the Divergence Test is the easiest way to show this. – user84413 Aug 14 '14 at 23:35
  • @IanMateus So you can prove clearly that the series is divergent, but it still converges to $-\frac{1}{12}$? – Aaron Wild Aug 14 '14 at 23:37
  • To make the situation clearer, you probably know $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}.$$ This means $$\sum_{k=0}^\infty 2^k=1+2+4+8+\cdots = \frac{1}{1-2}=-1,$$ right? Err, no. This does not converge to $-1$, we are outside the area of convergence $|x|<1$. But I can abuse the notation and write $1+2+4+\cdots=-1$ if I want, even though this is outside the area of convergence. This is very, very, similar to what is happening: the $-1$ is outside the area of convergence of the sum. I suggest you take a look at the many questions previously asked and try to digest the information. – Ian Mateus Aug 14 '14 at 23:41

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I think a requirement of the integral test is that $ f (n) $ be monotone decreasing.

Check http://www.math.com/tables/expansion/tests.htm. Under integral test.

dylan7
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