$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ ... + \frac{1}{1+2+3+...+n} = ?$ How do I simplify the following series
$$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$
From what I see, each term is the inverse of the sum of $n$ natural numbers.
Assuming there are $N$ terms in the given series,
$$a_N = \frac{1}{\sum\limits_{n = 1}^{N} n} = \frac{2}{n(n+1)}$$
$$\Rightarrow \sum\limits_{n=1}^{N} a_N = \sum\limits_{n=1}^{N} \frac{2}{n(n+1)}$$
... and I'm stuck.
I've never actually done this kind of problem before (am new to sequences & series). 
So, a clear and detailed explanation of how to go about it would be most appreciated.
PS- I do not know how to do a telescoping series!!

HINT: $$\frac {2} {n(n+1)} = \frac 2 n - \frac 2 {n+1}$$
