Question

If x is positive, the sum to infinity of the series
$\frac{1}{1+x}-\frac{1-x}{(1+x{)}^2}+\frac{(1-x{)}^2}{(1+x{)}^3}-\frac{(1-x{)}^3}{(1+x{)}^4}+......\mathrm{is}$
(a) 1/2
(b) 3/4
(c) 1
(d) none of these

Answer 1

(a) $\frac{1}{2}$

$$\begin{gathered} \mathrm{Let}S=\frac{1}{\left(1+x\right)}-\frac{\left(1-x\right)}{{\left(1+x\right)}^2}+\frac{{\left(1-x\right)}^2}{{\left(1+x\right)}^3}-\frac{{\left(1-x\right)}^3}{{\left(1+x\right)}^4}+...\infty \\ \mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{G}.\mathrm{P}.\mathrm{with}a=\frac{1}{\left(1+x\right)}\mathrm{and}r=-\frac{\left(1-x\right)}{\left(1+x\right)}. \\ \therefore S=\frac{a}{\left(1-r\right)} \\ \Rightarrow S=\frac{\frac{1}{\left(1+x\right)}}{\left[1-\left(-\frac{\left(1-x\right)}{\left(1+x\right)}\right)\right]} \\ \Rightarrow S=\frac{\frac{1}{\left(1+x\right)}}{\left[1+\frac{\left(1-x\right)}{\left(1+x\right)}\right]} \\ \Rightarrow S=\frac{\frac{1}{\left(1+x\right)}}{\left[\frac{\left(1+x\right)+\left(1-x\right)}{\left(1+x\right)}\right]} \\ \Rightarrow S=\frac{1}{2} \end{gathered}$$