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My friend asked me if $\displaystyle \lim_{n \to \infty}\frac{2}{n(n+1)}=-12$ and burst into laughter.

I'm a little bit confused because I thought the limit is $0$.

Is this a joke or something?

Saketh Malyala
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    See this page about why in some contexts it is sometimes correct to say "$1+2+3+4+\dots=-\frac{1}{12}$" and remember that $1+2+3+\dots+n=\frac{n(n+1)}{2}$ for finite values of $n$. That being said, without additional context, the more "obvious" interpretation is usually the agreed one and the limit is indeed zero. – JMoravitz Sep 07 '17 at 04:40

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In some very interesting and deep but intuitively nonsensical ways (and incorrect, according to the standard definition of infinite sums), we have that $\sum_{k=1}^\infty k = \frac{-1}{12}$.

Now just recall that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and apply limit rules using the above "equality" to get your friend's answer.

Brevan Ellefsen
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That equation, while absurd, is suggestive of the equation $$\sum_{n=1}^\infty n=-\frac{1}{12}$$ ...which is also absurd, because the limit of partial sums does not exist. On the other hand, the formal series $\sum_{n=1}^\infty n$ has a zeta-regularized sum, and a cutoff-regularized sum, and a Ramanujan sum, of $-\frac{1}{12}$. As the comments say, refer to Wikipedia, and also math.SE: Why does $1+2+3+\cdots = -\frac{1}{12}$?

So while the latter equation is not literally true, given the usual interpretation of its constituent symbols, it is suggestive of a true (and meaningful, and useful, and beautiful) result, just by replacing the usual sum $\sum$ with any of several generalized operators. Do keep in mind that evaluating sequential limits in the order topology on the real numbers is not the ultimate end of analysis.

By contrast, your friend's equation $\lim_{n\to\infty}\frac{2}{n(n+1)}=-12$ is not only false, but there is no evident way to generalize any part of it and get a true statement. So really, the joke is on your friend for not appreciating the difference. You are correct in that $\lim_{n\to\infty}\frac{2}{n(n+1)}=0$.

Chris Culter
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    Recall that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ so that one would naturally think $\frac{-1}{12} = \sum_{k=1}^\infty k = \lim_{n \to \infty} \frac{n(n+1)}{2}$. Now flip each of these by limit laws and you get the friend's result. Not that these steps are very justified at all, but the friend was joking :) – Brevan Ellefsen Sep 07 '17 at 04:59
  • @BrevanEllefsen Thanks, I could have been more explicit about the connection in my answer. My real point is that there's an asymmetry between the two equations. Starting from $\sum n$, there is a natural way to get to $-\frac{1}{12}$. But starting from $\lim\frac{2}{n(n+1)}$, there is no natural way to get to $-12$, nor is there any good reason to flip it and transform it into a sum. – Chris Culter Sep 07 '17 at 05:07
  • I agree completely. I just thought it was worth mentioning to show where the OP's friend's joke came from, even if the reasoning isnt the best. – Brevan Ellefsen Sep 07 '17 at 05:10
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HINT: Remember that stupid joke that $1+2+3....$ "tends" towards $-\frac{1}{12}$