Question

The sum of the series, $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+.....$ upto infinite terms. If $|x|<1$, is

• A. $\frac{x}{1-x}$
• B. $\frac{1}{1-x}$
• C. $\frac{1+x}{1-x}$
• D. 1

Answer 1

Answer: (A)

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

Answer 2

Let the given series be $S$. The general term of the series is $T_n = \frac{x^{2^{n-1}}}{1-x^{2^n}}$.

We use the identity: $\frac{A}{1-A^2} = \frac{1}{1-A} - \frac{1}{1-A^2}$.

Applying this identity with $A = x^{2^{n-1}}$, we get: $T_n = \frac{1}{1-x^{2^{n-1}}} - \frac{1}{1-x^{2^n}}$.

Now, we write out the first few terms of the series: $T_1 = \frac{x}{1-x^2} = \frac{1}{1-x} - \frac{1}{1-x^2}$ $T_2 = \frac{x^2}{1-x^4} = \frac{1}{1-x^2} - \frac{1}{1-x^4}$ $T_3 = \frac{x^4}{1-x^8} = \frac{1}{1-x^4} - \frac{1}{1-x^8}$ ...

The sum of the series is $S = T_1 + T_2 + T_3 + \dots$. This is a telescoping series.

The sum of the first $N$ terms, $S_N$, is: $S_N = \left(\frac{1}{1-x} - \frac{1}{1-x^2}\right) + \left(\frac{1}{1-x^2} - \frac{1}{1-x^4}\right) + \dots + \left(\frac{1}{1-x^{2^{N-1}}} - \frac{1}{1-x^{2^N}}\right)$ $S_N = \frac{1}{1-x} - \frac{1}{1-x^{2^N}}$.

To find the sum to infinite terms, we take the limit as $N \to \infty$.

Given $|x|<1$, as $N \to \infty$, $x^{2^N} \to 0$.

Therefore, $S = \lim_{N \to \infty} S_N = \frac{1}{1-x} - \frac{1}{1-0} = \frac{1}{1-x} - 1$.

$S = \frac{1 - (1-x)}{1-x} = \frac{x}{1-x}$.