Sequences & SeriesmediumFree

Each term splits as 1/(1-x^(2^(n-1))) - 1/(1-x^(2^n)), telescoping to 1/(1-x) - 1 = x/(1-x) for |x| < 1

JEE Maths question with a full step-by-step solution.

Question
The sum of the series x1−x2+x21−x4+x41−x8+⋯\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots to infinity, for ∣x∣<1|x|<1, is
Ax1−x\frac{x}{1-x}correct
B11−x\frac{1}{1-x}
C1+x1−x\frac{1+x}{1-x}
D11
Solution
Step 1: General term tn=x2n−11−x2nt_n=\dfrac{x^{2^{n-1}}}{1-x^{2^n}}. Write the numerator as (1+x2n−1)−1\big(1+x^{2^{n-1}}\big)-1 and factor 1−x2n=(1+x2n−1)(1−x2n−1)1-x^{2^n}=\big(1+x^{2^{n-1}}\big)\big(1-x^{2^{n-1}}\big):
tn=11−x2n−1−11−x2n.t_n=\frac{1}{1-x^{2^{n-1}}}-\frac{1}{1-x^{2^n}}.
 Step 2: This telescopes:
Sn=11−x−11−x2n.S_n=\frac{1}{1-x}-\frac{1}{1-x^{2^n}}.
 Step 3: For ∣x∣<1|x|<1, x2n→0x^{2^n}\to 0, so
S∞=11−x−1=x1−x.S_\infty=\frac{1}{1-x}-1=\frac{x}{1-x}.
 Correct answer: (1)
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