Question

For $x>0$ the sum of the series $\frac{1}{1+x}-\frac{(1-x)}{(1+x{)}^2}+\frac{(1+x{)}^2}{(1+x{)}^3}-.........\infty$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $1$

Answer 1

The correct option is B $\frac{1}{2}$

$\frac{1}{1+x}-\frac{(1-x)}{(1+x{)}^2}+\frac{(1-x{)}^2}{(1+x{)}^3}-\infty$

G.P.,

$a=\frac{1}{(1+x)},r=\frac{-(1-x)}{(1+x)}=\frac{x-1}{(x+1)}$

$S_{\infty }=\frac{a}{1-r}$

$=\frac{\frac{1}{(1+x)}}{1-\left(\frac{x-1}{x+1}\right)}=\frac{1}{2}$