Ltn→∞⁡1+a+a2+….+an−11+b+b2+….+bn−1=0 if

Ltn→∞1+a+⋅⋅⋅⋅+an−11+b+⋅⋅⋅⋅+bn−1=Ltn→∞an−1/a−1bn−1/b−1=Ltn→∞b−1a−1⋅an−1bn−1=Ltn→∞b−1a−1a/bn−1/21−1/bn=0 ifab 1.

Ltn→∞⁡1+a+a2+….+an−11+b+b2+….+bn−1=0 if

Ltn→∞1+a+⋅⋅⋅⋅+an−11+b+⋅⋅⋅⋅+bn−1=Ltn→∞an−1/a−1bn−1/b−1=Ltn→∞b−1a−1⋅an−1bn−1=Ltn→∞b−1a−1a/bn−1/21−1/bn=0 ifab 1.

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${Lt}_{n \to \infty} \frac{1 + a + a^{2} + \dots . + a^{n - 1}}{1 + b + b^{2} + \dots . + b^{n - 1}} = 0 if$

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$a = b$

$a < b$

$a > b$

$1 < a < b$

answer is B.

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Detailed Solution

$\begin{array}{l} \underset{n \to \infty}{L t} \frac{1 + a + \cdot \cdot \cdot \cdot + a^{n - 1}}{1 + b + \cdot \cdot \cdot \cdot + b^{n - 1}} = \underset{n \to \infty}{L t} \frac{(a^{n} - 1) / (a - 1)}{(b^{n} - 1) / (b - 1)} \\ = \underset{n \to \infty}{L t} \frac{b - 1}{a - 1} \cdot \frac{a^{n} - 1}{b^{n} - 1} \\ = \underset{n \to \infty}{L t} \frac{b - 1}{a - 1} \frac{{(a / b)}^{n} - 1 / 2}{1 - (1 / b^{n})} \\ = 0 i f \frac{a}{b} < 1, \frac{1}{b} < 1 \Rightarrow a < b, b > 1. \end{array}$