Series...

• Feb 15th 2010, 08:19 AM
bearej50
Series...
Find the sum of:

1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2))
• Feb 15th 2010, 09:17 AM
tonio
Quote:

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

I think you can take it from here (Wink)

Tonio
• Feb 15th 2010, 10:44 AM
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 $\displaystyle \frac{1}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{n+2-n}{n(n+1)(n+2)}=\frac{1}{2}\left( \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} \right),$ and this one telescopes easily.
• Feb 15th 2010, 06:45 PM
tonio
Quote:

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 $\displaystyle \frac{1}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{n+2-n}{n(n+1)(n+2)}=\frac{1}{2}\left( \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} \right),$ and this one telescopes easily.

Much better than my solution.

Tonio