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 $\infty$

• A. $\frac{x}{1-x}$
  
• B. $\frac{2x}{x-1}$
• C. $\frac{x}{x+1}$
• D. $Noneofthese$

Answer 1

Answer: (A)

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

If we combine the first $2$ terms, we get

$\frac{x(1+x^2)}{1-x^4}+\frac{x^2}{1-x^4}=\frac{x+x^2+x^3}{1-x^4}$

If we combine the first $3$ terms, we get

$\frac{(x+x^2+x^3)(1+x^4)}{1-x^8}+\frac{x^4}{1-x^8}$

$=\frac{x+x^2+x^3+x^4+x^5+x^6+x^7}{1-x^8}$

Following this pattern, you will find out that the sum of the first n terms is,

$\frac{x}{1-x^{2n}}\sum _{i=0}^{2^n-2}{x}^i$

And we should be able to evaluate this sum, because it is the finite sum of a geometric progression. Taking the limit $n\to \infty$ should yield the final answer of $\frac{x}{1-x}$.