Why is this sum what it is? I am trying to prove that
$$\frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+3+\cdots+n} = \frac{2n}{n+1}$$
I know the denominators have a closed form $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$ implying 
$$\frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+3+\cdots+n} = 2\sum_{k=1}^{n}\frac{1}{k(k+1)}$$
Can't figure out where to go from here.

To continue sheol's answer.
A telescope sum cancel out.
$2\sum_{k=1}^{n}\frac{1}{k(k+1)} = 2( \sum_{k=1}^{n}(\frac{1}{k} - \frac{1}{k+1}))$
$= 2(\sum_{k=1}^{n}\frac{1}{k} - \sum_{k=1}^{n}\frac{1}{k + 1})$
$= 2(\sum_{k=0}^{n-1}\frac{1}{k+1} - \sum_{k=1}^{n}\frac{1}{k + 1})$
$= 2(\frac 1{0 + 1} + \sum_{k=1}^{n-1}\frac{1}{k + 1} - \sum_{k=1}^{n-1}\frac{1}{k + 1} + \frac 1{n+1})$
