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 .....