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

$\frac{x}{1-x^2}+\frac{x^2}{1-x^2}+\frac{x^4}{1-x^4}........+\frac{x^{2n}}{1-x^{2n+1}}$

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

$=\frac{1}{1-x}-\frac{1}{1-x^{2n+1}}\to \frac{1}{1-x}-1$

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