Question

Sum of the series $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-....\infty$

• A. ${\log }_{e}2-1$
• B. ${\log }_{e}2$
• C. ${\log }_e\left(4/e\right)$
• D. $2{\log }_{e}2$

Answer 1

The correct option is C ${\log }_e\left(4/e\right)$

$S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-...\infty$
$S_1=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}-...\infty$
(sum of positive terms)
$T_n=\frac{1}{(2n-1)\left(2n\right)}=\frac{1}{2n-1}-\frac{1}{2n}$
$S_1=\sum T_n=\sum (\frac{1}{2n-1}-\frac{1}{2n})$
$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...\infty$
$={\log }_{e}2$ (A)
Again $S_2=\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...\infty$
(sum of negative terms)
$T_n{}^{\prime }=\frac{1}{\left(2n\right)(2n+1)}$
$S_2=\sum T_n{}^{\prime }=\sum \frac{1}{\left(2n\right)(2n+1)}=\sum (\frac{1}{2n}-\frac{1}{2n+1})$
$=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...\infty$
$=-[-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+...\infty ]$
$=-[{\log }_{e}2-1]=1-{\log }_{e}2$ ......(B)
Now $S=S_1-S_2$ $\left(A\right)-\left(B\right)$
${\log }_{e}2-1+{\log }_{e}2$
$={\log }_e\left(4/e\right)$