Question

The sum of $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+.....$ is

• A. $\ln \left(2e\right)$
• B. $\ln \left(4e\right)$
• C. $\ln \left(\frac{2}{e}\right)$
• D. $\ln \left(\frac{4}{e}\right)$

Answer 1

Answer: (D)

$\ln \left(\frac{4}{e}\right)$

As we know that, $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$
Put $x=1$
$\log 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...={\sum }_{n=1}^{\infty }\frac{{(-1)}^{n+1}}{n}=1+{\sum }_{n=1}^{\infty }\frac{{(-1)}^n}{n+1}$ ...(1)
$S_{\infty }=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+.....$

$\Rightarrow S_{\infty }={\sum }_{n=1}^{\infty }\frac{{(-1)}^{n+1}}{n(n+1)}={\sum }_{n=1}^{\infty }{(-1)}^{n+1}(\frac{1}{n}-\frac{1}{n+1})$

$\Rightarrow S_{\infty }={\sum }_{n=1}^{\infty }\frac{{(-1)}^{n+1}}{n}+{\sum }_{n=1}^{\infty }\frac{{(-1)}^n}{n+1}$
$\Rightarrow S_{\infty }=\log 2+\log 2-1=\log 2^2-\log e$ ...[ From (1) ]
$\Rightarrow S_{\infty }=\log \frac{4}{e}$
Ans: D