Question:
If x is positive, the sum to infinity of the series
$\frac{1}{1+x}-\frac{1-x}{(1+x)^{2}}+\frac{(1-x)^{2}}{(1+x)^{3}}-\frac{(1-x)^{3}}{(1+x)^{4}}+\ldots$ is
(a) 1/2
(b) 3/4
(c) 1
(d) none of these
Solution:
(a) $\frac{1}{2}$
Let $S=\frac{1}{(1+x)}-\frac{(1-x)}{(1+x)^{2}}+\frac{(1-x)^{2}}{(1+x)^{3}}-\frac{(1-x)^{3}}{(1+x)^{4}}+\ldots \infty$
It is clear that it is a G.P. with $a=\frac{1}{(1+x)}$ and $r=-\frac{(1-x)}{(1+x)}$.
$\therefore S=\frac{a}{(1-r)}$
$\Rightarrow S=\frac{\frac{1}{(1+x)}}{\left[1-\left(-\frac{(1-x)}{(1+x)}\right)\right]}$
$\Rightarrow S=\frac{\frac{1}{(1+x)}}{\left[1+\frac{(1-x)}{(1+x)}\right]}$
$\Rightarrow S=\frac{\frac{1}{(1+x)}}{\left[\frac{(1+x)+(1-x)}{(1+x)}\right]}$
$\Rightarrow S=\frac{1}{2}$