Question

${lim}_{x\to \infty }(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^n})$

Answer 1

${lim}_{x\to \infty }(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^n})$ ...(i)

This is G.P. of common ration,$r=\frac{1}{3}$

$\therefore$ Sum of terms of G.P with $a=\frac{1}{3},r=\frac{1}{3}$

$S_n=a\left(\frac{1-r^n}{1-r}\right)$

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

$=\frac{1}{3}\left(\frac{1-\frac{1}{3^n}}{\frac{2}{3}}\right)=\frac{1}{3}\times \frac{2}{3}(1-\frac{1}{3^n})$

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

Substituitng value of $S_n$ in (i),we get

${lim}_{n\to \infty }{S}_n={lim}_{n\to \infty }\frac{1}{2}(1-\frac{1}{3^n})$

$={lim}_{n\to \infty }\frac{1}{2}(1-\frac{1}{3^n})$

$=\frac{1}{2}(1-0)$

$=\frac{1}{2}$