If x is positive, the sum to infinity of the series

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

$\therefore S=\frac{a}{(1-r)}$

$\Rightarrow S=\frac{\frac{1}{(1+x)}}{\left[1-\left(-\frac{(1-x)}{(1+x)}\right)\right]}$

$\Rightarrow S=\frac{\frac{1}{(1+x)}}{\left[1+\frac{(1-x)}{(1+x)}\right]}$

$\Rightarrow S=\frac{\frac{1}{(1+x)}}{\left[\frac{(1+x)+(1-x)}{(1+x)}\right]}$

$\Rightarrow S=\frac{1}{2}$

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