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Q.
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
Sequences and Series
Solution:
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,
$S_{n}=\sum\limits _{n=1}^{n} t_{n}$
$\left[\left\{\frac{1}{1-x}-\frac{1}{1-x^{2}}\right\}+\left\{\frac{1}{1-x^{2}}-\frac{1}{1-x^{4}}\right\}+\ldots+\left\{\frac{1}{1-x^{2^{n-1}}}-\frac{1}{1-x^{2^n}}\right\}\right]$
$=\frac{1}{1-x}-\frac{1}{1-x^{2^n}}$
Therefore, the sum to infinite terms is
$\lim\limits_{n\to\infty} S_{n}=\frac{1}{1-x}-1$
$=\frac{x}{1-x} \, \left[\because \lim\limits_{n\to\infty} x^{2n}=0, as\left|x\right|<1\right]$