Find the sum of an infinite series if it converges

Question

How would I go about finding the sum of the following infinite series?

$\sum\limits_{n=1}^{\infty} \frac{1}{1+2+3+4+..+n}$

I'm assuming a limit of some sort. I apologize if this is trivial, I am not that great at math.

Answer 1

${1\over{n(n+1)}}={1\over n}-{1\over{n+1}}$

$S_n=2-{1\over {n+1}}$

Answer 2

From all the information in the comments:

$$\sum\limits_{n=1}^{\infty} \frac{1}{1+2+3+4+\cdots+n}$$ $$= \sum\limits_{n=1}^{\infty} \frac{1}{\frac{1}{2}n(n+1)}$$ $$=2 \sum\limits_{n=1}^{\infty} \frac{1}{n(n+1)}$$ $$=2 \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$

and now you can use telescoping series to finish.

Answer 3

This is equal to $$2\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$$ because $\sum_{i=1}^n i = \frac{i(i+1)}{2}$.

$\frac{1}{n(n+1)}$ can be written as its partial fraction decomposition $\frac{a}{n} + \frac{b}{n+1}$, such that $an+a+bn = 1$. Let $a=1$ and $b=-1$. $$2\sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+1} = 2(1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} \ldots) = 2(1) = 2$$ (This is a telescoping series)