Question

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

• A. 4/3
• B. 2/3
• C. 1/3
• D.

Answer 1

The correct option is A

4/3

$\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^n}}{1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^n}}$
Using the formula for finding the sum of n terms of G.P -
=$\underset{n\to \infty }{lim}\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}\times \frac{1-\frac{1}{3}}{1-\frac{1}{3^{n+1}}}[\therefore x^{n+1}\to 0asn\to \infty if-1<x<1]=\frac{4}{3}$