Question

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

• A. $2\log {}_{e}2+1$
• B. $2\log {}_{e}2$
• C. $2\log {}_{e}2-1$
• D. $\log {}_{e}2-1$

Answer 1

The correct option is C $2\log {}_{e}2-1$

$S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-.....\infty$

Let $S_1=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+......\infty$

$\therefore t_n=\frac{1}{(2n-1)\left(2n\right)}=\frac{1}{2n-1}-\frac{1}{2n}$

$\therefore S_n=\sum t_n=\sum (\frac{1}{2n-1}-\frac{1}{2n})$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}.....\infty$

$={\log }_{e}2----\left(1\right)$

Let $S_2=\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+....\infty$

$t_n^{\prime }=\frac{1}{\left(2n\right)(2n+1)}=$

$S_2=\sum t_n^{\prime }=\sum (\frac{1}{2n}-\frac{1}{2n+1})$

$=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+.....\infty$

$=-[-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+....\infty ]$

$=-[{\log }_{e}2-1]=1-{\log }_{e}2$-------$\left(2\right)$

Now $S=S_1-S_2$

$=\left(1\right)-\left(2\right)$

$S={\log }_{e}2-1+{\log }_{e}2=2{\log }_{e}2-1$