series problem .

Hello thereddevils

Quote:

Originally Posted by thereddevils

Find the sum to the n terms of the series :

$\frac{1}{(2r-1)(2r+1)(2r+3)}$

Using Partial Fractions, you find that

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

So if the sum to n terms is $S$ :

$8S = \sum_{r=1}^n\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)$

$= \sum_{r=1}^n\frac{1}{2r-1} -\sum_{r=2}^{n+1}\frac{2}{2r-1}+\sum_{r=3}^{n+2}\frac{1}{2r-1}$

$= 1 + \frac{1}{3}+\sum_{r=3}^n\frac{1}{2r-1}-(\frac{2}{3}+2\sum_{r=3}^n\frac{1}{2r-1} + \frac{2}{2n+1})+\sum_{r=3}^n\frac{1}{2r-1}+\frac{1}{2n+1}+\frac{1}{2n+3}$

$=\frac{2}{3}-\frac{2}{2n+1}+\frac{1}{2n+1}+\frac{1}{2n+3}$

$\Rightarrow S=\frac{1}{8}\left(\frac{2}{3}-\frac{1}{2n+1}+\frac{1}{2n+3}\right)$

Grandad