Sigma notation

• Jan 23rd 2008, 01:01 AM
SengNee
Sigma notation
$\displaystyle \sum_{r=1}^{n}\frac {2r}{(2r-1)(2r+1)(2r+3)}$

No need show me the method, just give me the answer, thanks.
• Jan 23rd 2008, 03:14 AM
mr fantastic
Quote:

Originally Posted by SengNee
$\displaystyle \sum_{r=1}^{n}\frac {2r}{(2r-1)(2r+1)(2r+3)}$

No need show me the method, just give me the answer, thanks.

From the usual partial fraction decomposition:

$\displaystyle \frac {2r}{(2r-1)(2r+1)(2r+3)} = \frac{1}{8} \left( \frac{1}{2r - 1}\right) + \frac{1}{4} \left( \frac{1}{2r + 1}\right) - \frac{3}{8} \left( \frac{1}{2r + 3}\right)$.

So the for n > 2, the first few terms of the series are:

$\displaystyle \frac{1}{8} \left( 1 + \frac{1}{3} + \frac{1}{5} + ....\right) + \frac{1}{4} \left( \frac{1}{3} + \frac{1}{5} + .....\right) - \frac{3}{8} \left( \frac{1}{5} + \frac{1}{7} + ....\right)$.

Note that $\displaystyle \frac{1}{8} + \frac{1}{4} - \frac{3}{8} = 0$ ......

So what terms will be left if n = 3? n = 4? n = n?

If I was less lazy I'd see this through to the answer, which I would then give you .....
• Jan 23rd 2008, 09:21 AM
topsquark
Quote:

Originally Posted by SengNee
No need show me the method, just give me the answer, thanks.

As this is a Math *HELP* Forum, I find this comment to be amazing. Why would you just want the answer and not learn how to do it?

-Dan
• Jan 23rd 2008, 06:41 PM
SengNee
Quote:

Originally Posted by mr fantastic
From the usual partial fraction decomposition:

$\displaystyle \frac {2r}{(2r-1)(2r+1)(2r+3)} = \frac{1}{8} \left( \frac{1}{2r - 1}\right) + \frac{1}{4} \left( \frac{1}{2r + 1}\right) - \frac{3}{8} \left( \frac{1}{2r + 3}\right)$.

So the for n > 2, the first few terms of the series are:

$\displaystyle \frac{1}{8} \left( 1 + \frac{1}{3} + \frac{1}{5} + ....\right) + \frac{1}{4} \left( \frac{1}{3} + \frac{1}{5} + .....\right) - \frac{3}{8} \left( \frac{1}{5} + \frac{1}{7} + ....\right)$.

Note that $\displaystyle \frac{1}{8} + \frac{1}{4} - \frac{3}{8} = 0$ ......

So what terms will be left if n = 3? n = 4? n = n?

If I was less lazy I'd see this through to the answer, which I would then give you .....

n=n
• Jan 23rd 2008, 06:43 PM
SengNee
Quote:

Originally Posted by topsquark
As this is a Math *HELP* Forum, I find this comment to be amazing. Why would you just want the answer and not learn how to do it?

-Dan

Because I had know how to do and I did it step by step, but my answer is different to the book's answer.
• Jan 24th 2008, 02:30 AM
topsquark
Quote:

Originally Posted by SengNee
Because I had know how to do and I did it step by step, but my answer is different to the book's answer.

Then it's probably best to at least post your solution to see who agrees with it.

-Dan