Question

Sum the following series to n terms and to infinity $\frac{1}{\mathrm{1.3.5}}+\frac{1}{\mathrm{3.5.7}}+\frac{1}{\mathrm{5.7.9}}+....$.

Answer 1

General term($n^{th}$ term) $=\frac{1}{(2n-1)(2n+1)(2n+3)}$

By partial fraction we can write it

as $\frac{1}{8}[\frac{1}{(2n-1)}-\frac{2}{(2n+1)}+\frac{1}{(2n+3)}]$ ......... $\left(i\right)$

Solving for $\sum _{n=1}^n[\frac{1}{2n-1}-\frac{1}{2n+1}]$

$=(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7}).....(\frac{1}{2n-1}-\frac{1}{2n+1})$

$=1-\frac{1}{2n+1}$ ...... $\left(ii\right)$

Similarly for ${\sum }_{n=1}^n[\frac{1}{2n+1}-\frac{1}{2n+3}]$

We will get

$=\frac{1}{3}-\frac{1}{2n+3}$ ...... $\left(iii\right)$

So from equation $\left(ii\right)$ and $\left(iii\right)$ putting in eq $\left(i\right)$

We get

$\frac{1}{8}[\frac{1}{(2n-1)}-\frac{2}{(2n+1)}+\frac{1}{(2n+3)}]$

$=\frac{1}{8}\{[1-\frac{1}{2n+1}]-[\frac{1}{3}-\frac{1}{2n+3}]\}$

$=\frac{1}{4}[\frac{1}{3}-\frac{1}{(2n+1)(2n+3)}]$

Sum to infinity ($n\to \infty$)

$=\frac{1}{4}[\frac{1}{3}-0]$

$=\frac{1}{12}$