Here in example 4 this result is shown.
${s_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{{3^{i - 1}}}}} = \dfrac{3}{2}\left( {1 - \dfrac{1}{{{3^n}}}} \right)$
Now without looking at its solution I was trying like this:
${s_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{{3^{i - 1}}}}}=3\sum\limits_{i = 0}^n {\dfrac{1}{{{3^{i }}}}}=3\bigg(\dfrac{1-(\frac{1}{3})^{n+1}}{1-\frac{1}{3}}\bigg)=\mathop {\lim }\limits_{n \to \infty }\dfrac{3}{2}\cdot3(1-(\frac{1}{3})^{n+1})=\dfrac{9}{4}$
What am I doing wrong why I am not getting answer $\dfrac{3}{2}$
$S_n=\sum_{i=1}^n \frac{1}{3^{i-1}}=1+\frac 13+...+\frac{1}{3^{n-1}}=\frac{1-\frac{1}{3^n}}{1-\frac13}=\frac 32(1-\frac{1}{3^n})$ Thus $\lim_{n \to\infty}S_n=\frac 32$. If you are considering the infinite geometric series with $|r|<1$ then sum=$\frac{a}{1-r}$ where $a$ is the first term of the series. Here $a=1,r=\frac 13$. So sum is $\frac{1}{1-\frac 13}=\frac 32$