<s>[INST] 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   
 [/INST] Your work so far is good. Just keep going.
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{2} \left(\frac{1}{1 \cdot 3 \cdot 5 \cdots (2n-1)} - \frac{1}{1 \cdot 3 \cdot 5 \cdots (2n+1)}\right)
= \frac{1}{2} \left(\frac{1}{1} - \frac{1}{1 \cdot 3} + \frac{1}{1\cdot 3} - \frac{1}{1 \cdot 3 \cdot 5} + \frac{1}{1 \cdot 3 \cdot 5} - \frac{1}{1 \cdot 3 \cdot 5 \cdot 7} + \cdots\right)$$
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