Question

$\underset{n\to \infty }{\lim }\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\right]$

Answer 1

$$\begin{gathered} \underset{n\to \infty }{\lim }\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...\frac{1}{3^n}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{1}{3}\left(1+\frac{1}{3}+\frac{1}{3^2}+.....\frac{1}{3^{n-1}}\right)\right] \\ =\underset{n\to \infty }{\lim }\frac{\left[{\displaystyle \frac{1}{3}}\left(1-{\displaystyle \frac{1}{3^n}}\right)\right]}{1-{\displaystyle \frac{1}{3}}}\left[a+ar+a{r}^2+......a{r}^{n-1}=a\left(\frac{1-r^n}{1-r}\right)\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{{\displaystyle \frac{1}{3}}\left(1-{\displaystyle \frac{1}{3^n}}\right)}{{\displaystyle \frac{2}{3}}}\right] \\ =\underset{n\to \infty }{\lim }\frac{1}{2}\left[\left(1-\frac{1}{3^n}\right)\right] \\ \mathrm{As}n\to \infty ,3^n\to \infty ,\frac{1}{3^n}\to 0 \\ =\frac{1}{2}\left(1-0\right) \\ =\frac{1}{2} \end{gathered}$$