The least positive integer $\mathrm{n}$ such that $1-\frac{2}{3}-\frac{2}{3^2}-\ldots .-\frac{2}{3^{n-1}}

Question

The least positive integer $\mathrm{n}$ such that $1-\frac{2}{3}-\frac{2}{3^2}-\ldots .-\frac{2}{3^{n-1}} < \frac{1}{100}$, is:

Aicharya · Accepted Answer

$ \begin{aligned} &1-\frac{2}{3}-\frac{2}{3^2} \ldots \cdot \frac{2}{3^{n-1}} < \frac{1}{100} \ &\Rightarrow 1-\frac{2}{3}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \frac{1}{3^{n-1}}ight] < \frac{1}{100} \ &\Rightarrow \frac{1-2\left[\frac{1}{3}\left(\frac{1}{3^n}-1ight)ight]}{\frac{1}{3}-1} < \frac{1}{100} \ &\Rightarrow 1-2\left[\frac{3^n-1}{2.3^n}ight] < \frac{1}{100} \ &\Rightarrow 1-\left[\frac{3^n-1}{3^n}ight] < \frac{1}{100} \ &\Rightarrow 1-1+\frac{1}{3^n} < \frac{1}{100} \end{aligned} $ $ \Rightarrow 100 < 3^n $ Thus, least value of $n$ is 5

The least positive integer $\mathrm{n}$ such that $1-\frac{2}{3}-\frac{2}{3^2}-\ldots .-\frac{2}{3^{n-1}} <…

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