Question

Find $\underset{n\to 0}{lim}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n})$.

Answer 1

This is G.P. with first term, $a=\frac{1}{3};$

Common ratio $=\frac{1}{3}$

Sum of n terms of G.P. $=\frac{a(r^n-1)}{r-1}=\frac{\frac{1}{3}({\frac{1}{3}}^n-1)}{\frac{1}{3}-1}$

$S_n=-\frac{1}{2}(3^{-n}-1)$

$\underset{x\to 0}{lim}{S}_n=\frac{-1}{2}(1-1)=0$.