$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}$ I want to see OTHER approaches than this one. Make sure they are significantly different and not a direct restatement.
$$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}\tag{1}$$
$$\sum_{n=1}^x n=\frac{x(x+1)}{2} \; \forall x >0\tag{2}$$
$$\begin{align*}
(1)&\stackrel{(2)}{\iff} \frac{2}{2\cdot 3}+\frac{2}{3\cdot 4}+\frac{2}{4\cdot 5}+\dots+\frac{2}{x(x+1)}=\frac{2011}{2013}\\\\
&\iff 2\left (\frac12 -\frac13+\frac13-\frac14+\frac14-\dots+\frac{1}{x}-\frac{1}{x+1}\right )=\frac{2011}{2013}\\\\
&\iff 1-\frac{2}{x+1}=\frac{2011}{2013}\\\\
&\iff x=2012
\end{align*}$$

Let's prove by induction that
$$\sum_{m=2}^n \frac1{\sum_{k=1}^m k} = \frac{n-1}{n+1}$$
The base case $n=2$ is true. Now the induction step:
$$\sum_{m=2}^{n+1} \frac1{\sum_{k=1}^m k}=\sum_{m=2}^n \frac1{\sum_{k=1}^m k}+\frac1{\sum_{k=1}^{n+1}k}=\frac{n-1}{n+1}+\frac2{(n+1)(n+2)}=\frac{n^2+n}{(n+1)(n+2)}=\frac{n}{n+2}$$
Therefore the answer is $n=2012$.
(I know this is kind of lame, but the formula is easy to guess)
