Question

$\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^n}}{1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^n}}$ is equal to

• A. $\frac{4}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. none of these

Answer 1

The correct option is A $\frac{4}{3}$

$\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^n}}{1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^n}}$
$=\frac{\frac{1}{1-1/2}}{\frac{1}{1-1/3}}=\frac{4}{3}$
Hence, option 'A' is correct.