Question

The sum of infitnite series
$\frac{1}{\mathrm{1.2.3.4}}+\frac{4}{\mathrm{3.4.5.6}}+\frac{9}{\mathrm{5.6.7.8}}+\frac{16}{\mathrm{7.8.9.10}}+...$ is $\frac{1}{a}{\log }_{e}b-\frac{1}{c}$. Find $c-a\times b$
(Note: $a,b$ and $c$ are integers)

Answer 1

Let $T_n$ be the $n$th term of the given series. Then,
$T_n=\frac{n^2}{(2n-1)2n(2n+1)(2n+2)}$

$=\frac{n}{2(2n-1)(2n+1)(2n+2)}=\frac{A}{2n-1}+\frac{B}{2n+1}+\frac{C}{2n+2}$ (say)
$A=\underset{(2n-1)\to 0}{lim}(2n-1)\left[\frac{n}{2(2n-1)(2n+1)(2n+2)}\right]$
$=\underset{n\to \frac{1}{2}}{lim}\left[\frac{n}{2(2n+1)(2n+2)}\right]$
$=\frac{1}{24}$
$B=\underset{(2n+1)\to 0}{lim}(2n+1)\left[\frac{n}{2(2n-1)(2n+1)(2n+2)}\right]$
$=\underset{n\to -\frac{1}{2}}{lim}\left[\frac{n}{2(2n-1)(2n+2)}\right]$
$=\frac{1}{8}$
and $C=\underset{(2n+2)\to 0}{lim}(2n+2)\left[\frac{n}{2(2n-1)(2n+1)(2n+2)}\right]$
$=\underset{n\to -1}{lim}\left[\frac{n}{2(2n-1)(2n+1)}\right]$
$=-\frac{1}{6}$
$\therefore$ $T_n=\frac{1}{24}.\frac{1}{2n-1}+\frac{1}{8}.\frac{1}{2n+1}-\frac{1}{6}.\frac{1}{2n+2}$
$=\frac{1}{24}[\frac{1}{2n-1}+\frac{3}{2n+1}-\frac{4}{2n+2}]$
$=\frac{1}{24}[\frac{1}{2n-1}+\frac{4-1}{2n+1}-\frac{4}{2n+2}]$

$=\frac{1}{24}[(\frac{1}{2n-1}-\frac{1}{2n+1})+4(\frac{1}{2n+1}-\frac{1}{2n+2})]$
Substituting $n=1,2,3,4$, ... and adding, we get

$=\frac{1}{24}[(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...)+4(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...)]$
$=\frac{1}{24}[1+4((1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...)-(1-\frac{1}{2}))]$
$=\frac{1}{24}[1+4({\log }_{e}2-\frac{1}{2})]$
$=\frac{1}{24}[1+4{\log }_{e}2-2]$
$=\frac{1}{6}{\log }_{e}2-\frac{1}{24}$
$\therefore c-a\times b=24-6\times 2=12$