Question

Let the series be $\frac{1}{\mathrm{1.3.5}}+\frac{1}{\mathrm{3.5.7}}+\frac{1}{\mathrm{5.7.9}}+...$ Deduce the sum of the series upto infinity.

• A. $\frac{1}{9}$
• B. $\frac{1}{12}$
• C. $\frac{1}{16}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{1}{12}$

$T_n=\frac{1}{(2n-1)(2n+1)(2n+3)}$
Split into partial fractions by the method of suppression by putting $2n=1,-1,-3$ successively.
$\therefore T_n=\frac{1}{8(2n-1)}-\frac{2}{8(2n+1)}+\frac{1}{8(2n+3)}$
$\therefore T_n=\frac{1}{8}[(\frac{1}{2n-1}-\frac{1}{2n+1})-(\frac{1}{2n+1}-\frac{1}{2n+3})]$
Now putting $n=1,2,3,...n$ and adding
$S_n=\frac{1}{8}[(1-\frac{1}{2n+1})-(\frac{1}{3}-\frac{1}{2n+3})]$
$=\frac{1}{8}[\frac{2n}{2n+1}-\frac{2n}{3(2n+3)}]=\frac{n}{4}\left[\frac{4n+8}{3(2n+1)(2n+3)}\right]$
$=\frac{n(n+2)}{3(2n+1)(2n+3)}$
In order to find $S_{\infty }$ we re-write the above
$S_n=\frac{n^2(1+\frac{2}{n})}{3.4n^2(1+\frac{1}{2n})(1+\frac{3}{2n})}$
when $n\to \infty$
$S_n=\frac{1}{12}$