Question

The sum of the series $\left(\frac{1}{1·2}\right)-\left(\frac{1}{2·3}\right)+\left(\frac{1}{3·4}\right)-........$is

• A. $2\log \left(2\right)+1$
• B. $2\log \left(2\right)-1$
• C. $2\log \left(2\right)$
• D. $\log \left(2\right)-1$

Answer 1

The correct option is B

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

Explanation for correct options

Let $S=\left(\frac{1}{1·2}\right)-\left(\frac{1}{2·3}\right)+\left(\frac{1}{3·4}\right)-........$

$$\begin{gathered} S=\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)-................. \\ \Rightarrow S=1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+............ \\ \Rightarrow S=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+........\right)+\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........\right)-----------\left(1\right) \end{gathered}$$

$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+..........$

put $x=1$

$$\begin{gathered} \log \left(2\right)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....... \\ \Rightarrow \log \left(2\right)-1=-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....... \end{gathered}$$

equation $\left(1\right)$ becomes

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

Hence, $\mathrm{Option}\left(\mathrm{B}\right)$ is correct.