# Math Help - Series...

1. ## Series...

Find the sum of:

1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2))

2. Originally Posted by bearej50
Find the sum of:

1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2))

Telescopic series: $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\Longrightarrow \frac{1}{n(n+1)(n+2)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{1}{n+2}=$ $\frac{1}{n(n+2)}$ $-\frac{1}{(n+1)(n+2)}$ $=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$ .

I think you can take it from here

Tonio

3. note that in tonio's solution the first term telescopes not so fast than the second term which does faster.

another way on doing this is to see that $\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{n+2-n}{n(n+1)(n+2)}=\frac{1}{2}\left( \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} \right),$ and this one telescopes easily.

4. Originally Posted by Krizalid
note that in tonio's solution the first term telescopes not so fast than the second term which does faster.

another way on doing this is to see that $\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{n+2-n}{n(n+1)(n+2)}=\frac{1}{2}\left( \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} \right),$ and this one telescopes easily.

Much better than my solution.

Tonio