Weird Limit $\frac{2}{n(n+1)}$ 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? 
 A: 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.
A: 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 $\lim_{n\to\infty}\frac{2}{n(n+1)}=0$.
A: HINT: Remember that stupid joke that $1+2+3....$ "tends" towards $-\frac{1}{12}$
