Question

Find the sum to infinite terms of the series $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+.....$

Answer 1

We need to find value of $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+.....$

$=\frac{(1+x)-1}{(1+x)(1-x)}+\frac{(1+x^2)-1}{(1+x^2)(1-x^2)}+\frac{(1+x^4)-1}{(1+x^4)(1-x^4)}+.....$

$=(\frac{1}{1-x}-\frac{1}{1-x^2})+(\frac{1}{1-x^2}-\frac{1}{1-x^4})+(\frac{1}{1-x^4}-\frac{1}{1-x^8})+.....$

$=\frac{1}{1-x}+(\frac{1}{1-x^2}-\frac{1}{1-x^2})+(\frac{1}{1-x^4}-\frac{1}{1-x^4})+.....-\frac{1}{1-x^{\infty }}$

$=\frac{1}{1-x}-\frac{1}{1-x^{\infty }}$

$=\frac{1}{1-x}$

Therefore, the sum of the given infinite series is $\frac{1}{1-x}$.