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. $\frac{x}{1-x}$
• B. $\frac{1}{1-x}$
• C. $\frac{1+x}{1-x}$
• D. $1$

Answer 1

Answer: (A)

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