compute $\frac{1}{2}(\frac{1}{3.5........(2n-1)} -\frac{1}{3.5........(2n+1)})$ compute the summation
$\sum_ {n=1}^{\infty} \frac{n}{3.5........(2n+1)}=  ?$
My attempts  : i take  $a_n =\frac{n}{3.5........(2n+1)}$
Now =$\frac{1}{2}$ .$\frac{(2n+ 1) - 1}{3.5........(2n+1)}= \frac{1}{2}(\frac{1}{3.5........(2n-1)} -\frac{1}{3.5........(2n+1)})$
after  that  i can not  able  to proceed  further,,,
pliz help me..
thanks  u   

You don't need to compute $a_n = \frac{1}{2}(\frac{1}{3.5…(2n-1)} -\frac{1}{3.5…(2n+1)})$
You need to compute $\sum_{n=1}^K a_n$.
Which is $\frac 12(\frac 11 - \frac 13) + \frac 12(\frac 13-\frac 1{3\cdot 5})+.....+ \frac{1}{2}(\frac{1}{3.5…(2K-3)} -\frac{1}{3.5…(2K-1)}) + \frac{1}{2}(\frac{1}{3.5…(2K-1)} -\frac{1}{3.5…(2K+1)})$
And the compute $\sum_{n=1}^\infty a_n =\lim\limits_{K\to \infty}\sum_{n=1}^K a_n$
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Note:  If $a_n = \frac{1}{2}(\frac{1}{3.5…(2n-1)} -\frac{1}{3.5…(2n+1)})$ we can let $b_n =\frac{1}{3.5…(2n-1)}$ 
and then we have 
$$\sum_{n=1}^{\infty}\frac 12 (b_n - b_{n+1})$$.
And you should be able so solve that even if you don't have any frigging idea what $b_n$ was defined as.
