How do i find the value for the series $ \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$? I have a general problem understand how to solve these kind of questions where you have a series and you need to find the value of it:
$$  \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} $$
I know that $  \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} $ = $ \frac{a_1}{n}+\frac{a_2}{n+1}+\frac{a_3}{n+2} $ which can be rewritten as: $ n^2(a_1+a_2+a_3)+n^1(3a_1+2a_2+a_3)n^0(2a_1)=1$
Using a LGS i come up with the solutions:
$ a_1 = \frac{1}{2}$;
$ a_2 = -1 $;
$ a_3 = \frac{1}{2} $;
Okay, So i put that into my initial formula:
$  \sum_{n=1}^{\infty} \frac{0.5}{n}+\frac{-1}{n+1}+\frac{0.5}{n+2} $
Nice, but here i need a clear guidance whats next as this step is confusing me. It would be really great so see a solution for this so that i can study it further.
Thanks in advance everyone!

You came close. Once you render
$  \sum_{n=1}^{\infty} \frac{0.5}{n}+\frac{-1}{n+1}+\frac{0.5}{n+2} $
break the middle term into two:
$  \sum_{n=1}^{\infty} \frac{0.5}{n}+\frac{-0.5}{n+1}+\frac{-0.5}{n+1}+\frac{0.5}{n+2} $
$=0.5[\sum_{n=1}^{\infty} (\frac{1}{n}-\frac{1}{n+1})-\sum_{n=1}^{\infty} (\frac{1}{n+1}-\frac{1}{n+2})]$
where the sums at the end each have terms of the form $f(n)-f(n+1)$ and therefore telescope to give the desired result. Thus
$\sum_{n=1}^{\infty} (\frac{1}{n}-\frac{1}{n+1})=(1/1-1/2)+(1/2-1/3)+...=1/1=1$
$\sum_{n=1}^{\infty} (\frac{1}{n+1}-\frac{1}{n+2})=(1/2-1/3)+(1/3-1/4)+...=1/2$
and the entire sum is/then $(1/2)(1-1/2)=1/4$.
