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

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Solution

Let $s = \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots \dots + \frac{1}{3^{n}} + \dots \dots \frac{1}{\infty}$
So, $s = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}$
Using sum of $G P$ to $\infty$ .

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