Question

The sum of the series $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+......$ to infinite terms, if |x| > 1 is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The general term of the series is $t_n=\frac{x^{2^{n-1}}}{1-x^{2^n}}$
$=\frac{1+x^{2^{n-1}}-1}{(1+x^{2^{n-1}})(1-x^{2n-1})}$
$\therefore 1_n=\frac{1}{1-x^{2n-1}}-\frac{1}{1-x^{2^n}}$
Now, $S_n={\sum }_{n-1}^n{t}_n=\left[\{\frac{1}{1-x}-\frac{1}{1-x^2}\}\right]$
$=\{\frac{1}{1-x^2}-\frac{1}{1-x^4}\}+....$
$=\{\frac{1}{1-x^{2^{n-1}}}-\frac{1}{1-x^{2^n}}\}]=\frac{1}{1-x}-\frac{1}{1-x^{2^n}}$
$\therefore$ The sum to infinite terms
$={lim}_{n\to \infty }{S}_n=\frac{1}{1-x}-1=\frac{x}{1-x}$
$[\because {lim}_{n\to \infty }{x}^{2^n}=0,as|x|<1]$