Question

The sum of infitnite series
$\frac{1}{1.3}+\frac{1}{2.5}+\frac{1}{3.7}+\frac{1}{4.9}+...\infty$ is $c-{\log }_{e}d$. Find $c^2+d^3$
(Note : $c$ and $d$ are positive integers)

Answer 1

Let $T_n$ be the $n$th term of the given series. Then,
$T_n=\frac{1}{n(2n+1)}$
Let $\frac{1}{n(2n+1)}=\frac{A}{n}+\frac{B}{2n+1}$
$\therefore$ $A=\underset{n\to 0}{lim}n\left[\frac{1}{n(2n+1)}\right]$
$=\underset{n\to 0}{lim}\left[\frac{1}{2n+1}\right]=\frac{1}{0+1}=1$
and $B=\underset{2n+1\to 0}{lim}(2n+1)\left[\frac{1}{n(2n+1)}\right]$
$=\underset{n\to -\frac{1}{2}}{lim}\left[\frac{1}{n}\right]$
$=-2$
$\therefore$ $\frac{1}{n(2n+1)}=\frac{1}{n}-\frac{2}{2n+1}$
$\therefore$ $T_n=\frac{1}{n}-\frac{2}{2n+1}$
Hence the sum of the series
$S=\sum _{n=1}^{\infty }{T}_n=\sum _{n=1}^{\infty }\frac{1}{n}-\frac{2}{2n+1}$

$=(1-\frac{2}{3})+(\frac{1}{2}-\frac{2}{5})+(\frac{1}{3}-\frac{2}{7})+...$
$=(1+\frac{1}{2}+\frac{1}{3}+...)-2(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)$
$=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...$
$=2-(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+...)$
$=2-{\log }_{e}2$
$\therefore c^2+d^3=4+8=12$