$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ ... + \frac{1}{1+2+3+...+n} = ?$

Question

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!!

Answer 1

$$ \sum_{n = 1}^{N}{2 \over n\left(n + 1\right)} = 2\sum_{n = 1}^{N}\left({1 \over n} - {1 \over n + 1}\right) = 2\left[1 - {1 \over 2} + {1 \over 2} - {1 \over 3} + \cdots + {1 \over N} - {1 \over N + 1}\right] $$ $$ \sum_{n = 1}^{N}{2 \over n\left(n + 1\right)} = 2\left(1 - {1 \over N + 1}\right)\quad\to\quad \color{#0000ff}{\Large 2}\quad\mbox{when}\quad N \to \infty $$

Answer 2

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

Answer 3

Hint:

$$\frac{2}{n(n+1)} = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$$

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NOTE:

$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$ Hence: $$\frac{m}{n(n+1)} = m\left(\frac{1}{n} - \frac{1}{n+1}\right)$$