Question

If $S=\frac{5}{1\times 3\times 7}+\frac{7}{3\times 5\times 9}+\frac{9}{5\times 7\times 11}+...,$ then the value of $45S$ is

Answer 1

$S=\frac{5}{1\times 3\times 7}+\frac{7}{3\times 5\times 9}+\frac{9}{5\times 7\times 11}+...$

Numerator $=5,7,9$
$\Rightarrow T_n=2n+3,n=1,2,3,...$
Similarly,
Denominator $=(2n-1)(2n+1)(2n+5),n=1,2,3,...$

$\Rightarrow S=\sum _{n=1}^{\infty }\frac{2n+3}{(2n-1)(2n+1)(2n+5)}$

By partial fraction
$\frac{2n+3}{(2n-1)(2n+1)(2n+5)}=\frac{a}{2n-1}+\frac{b}{2n+1}+\frac{c}{2n+5}$
On comparing the coefficients, we get
$a=\frac{1}{3},b=-\frac{1}{4},c=-\frac{1}{12}$

$\therefore S=\sum _{n=1}^{\infty }(\frac{1}{3(2n-1)}-\frac{1}{4(2n+1)}-\frac{1}{12(2n+5)})$
$=\frac{1}{3}\sum _{n=1}^{\infty }\frac{1}{2n-1}-\frac{1}{4}\left(\sum _{n=1}^{\infty }\frac{1}{2n-1}-1\right)-\frac{1}{12}\left(\sum _{n=1}^{\infty }\frac{1}{2n-1}-1-\frac{1}{3}-\frac{1}{5}\right)$
$=(\frac{1}{3}-\frac{1}{4}-\frac{1}{12})\sum _{n=1}^{\infty }\frac{1}{2n-1}+\frac{1}{4}+\frac{1}{12}(1+\frac{1}{3}+\frac{1}{5})$
$=0\times \sum _{n=1}^{\infty }\frac{1}{2n-1}+\frac{1}{4}+\frac{1}{12}\left(\frac{23}{15}\right)$
$S=\frac{17}{45}$
$\Rightarrow 45S=17$


Note:
$\sum _{n=1}^{\infty }\frac{1}{2n-1}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+..$
$\sum _{n=1}^{\infty }\frac{1}{2n+1}=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+..$
$\sum _{n=1}^{\infty }\frac{1}{2n+5}=\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+..$

$\Rightarrow \sum _{n=1}^{\infty }\frac{1}{2n+1}=\sum _{n=1}^{\infty }\frac{1}{2n-1}-1$
$\Rightarrow \sum _{n=1}^{\infty }\frac{1}{2n+5}=\sum _{n=1}^{\infty }\frac{1}{2n-1}-1-\frac{1}{3}-\frac{1}{5}$