$\lim _{n \rightarrow \infty}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots . .+\frac{1}{3^n}\right]$ We know that sum of geometric series $=\frac{a\left(1-r^n\right)}{1-r}$ while $r<1$. So sum of given series is : $\begin{array}{l}\frac{\frac{1}{3}\left[1-\left(\frac{1}{3}\right)^n\right]}{1-\frac{1}{3}}=\frac{1-\left(\frac{1}{3}\right)^n}{2} \\\therefore \lim _{n \rightarrow \infty}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots ..+\frac{1}{3^n}\right]=\lim _{n \rightarrow \infty} \frac{\left(1-\left(\frac{1}{3}\right)^n\right)}{2} \\\frac{\left(1-\frac{1}{3^{\infty}}\right)}{2}=\frac{\left(1-\frac{1}{\infty}\right)}{2}=\frac{1-0}{2}=\frac{1}{2}\end{array}$
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