Mathematics - Sequence & Series Question with Solution | TestHub

The general term of the series is $t_n=\frac{x^{2^{n-1}}}{1-x^{2^n}}$

$=\frac{1+x^{2^{n-1}}-1}{\left(1+x^{2^{n-1}}\right)\left(1-x^{2n-1}\right)}$

$\therefore 1_n=\frac{1}{1-x^{2n-1}}-\frac{1}{1-x^{2^n}}$

Now, $S_n={\sum }_{n=1}^n{t}_n=\left[\left\{\frac{1}{1-x}-\frac{1}{1-x^2}\right\}\right]$ $+\left\{\frac{1}{1-x^2}-\frac{1}{1-x^4}\right\}+\dots$. $+\left\{\frac{1}{1-x^{2^{n-1}}}-\frac{1}{1-x^{2^n}}\right\}]=\frac{1}{1-x}-\frac{1}{1-x^{2^n}}$

∴ The sum to infinite terms

$={\lim }_{n\to \infty }{S}_n=\frac{1}{1-x}-1=\frac{x}{1-x}$

$[\because {\mathrm{Lim}}_{n\to \infty }{x}^{2^n}=0$, as $|x|<1]$