<s>[INST] $\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}$ I want to see OTHER approaches than this one. Make sure they are significantly different and not a direct restatement.
$$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}\tag{1}$$
$$\sum_{n=1}^x n=\frac{x(x+1)}{2} \; \forall x >0\tag{2}$$
$$\begin{align*}
(1)&\stackrel{(2)}{\iff} \frac{2}{2\cdot 3}+\frac{2}{3\cdot 4}+\frac{2}{4\cdot 5}+\dots+\frac{2}{x(x+1)}=\frac{2011}{2013}\\\\
&\iff 2\left (\frac12 -\frac13+\frac13-\frac14+\frac14-\dots+\frac{1}{x}-\frac{1}{x+1}\right )=\frac{2011}{2013}\\\\
&\iff 1-\frac{2}{x+1}=\frac{2011}{2013}\\\\
&\iff x=2012
\end{align*}$$
 [/INST] I'm not sure there is much which can be done radically differently. The proof can be recast as an induction - which is what you might do after computing the first few sums by hand as $\frac 13, \frac 12=\frac 24, \frac 35, \frac 23=\frac 46, \frac 57$
The $n^{th}$ term is $\cfrac 2{(n+1)(n+2)}$
We can prove that the sum of the first $n$ terms is $\cfrac n{n+2}$, since this is true for the first term and $$\frac n{n+2}+\frac 2{(n+2)(n+3)}=\frac {n+1}{n+3}$$
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