Question

Find the sum of the series $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+......$ upto $n$ term.

Answer 1

Given,

$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}......\infty ;\left|x\right|<1T_n=\frac{x^{2n-1}}{1-x^{2n}}=\frac{1+x^{2n-1}-1}{(1+x^{2n-1})(1-x^{2n-1})}=\frac{1}{1-x^{2n-1}}-\frac{1}{1-x^{2n}}{S}_n={\sum }_{n=1}^n{T}_n=\{\frac{1}{1-x}-\frac{1}{1-x^2}\}+\{\frac{1}{1-x^2}-\frac{1}{1-x^4}\}+......\{\frac{1}{1-x^{2n-1}}-\frac{1}{-x^{2n}}\}=\frac{1}{1-x}-\frac{1}{1-x^{2n}}{S}_{\infty }={lim}_{n\to \infty }{S}_n=\frac{1}{1-x}-\frac{1}{1-x^{2n}}=\frac{1}{1-x}-1(\because \left|x\right|<1;{lim}_{n\to \infty }{x}^{2n}=0)=\frac{x}{1-x}$