91Ó°ÊÓ

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One-nth of the contaminant is eliminated in the first step of a water purification procedure. Each succeeding stage removes one-tenth of the impurity that was eliminated in the preceding stage.

The expression for sum of infinite GP series.

$S=\frac{a}{1-r}$

Where, S is the sum of GP up to infinite terms.

Another geometric series-described method is water purification, as explained here. If each stage of purification removes one-nth of the current impurity, the total proportion of the impurity removed after N stages is given by

$\begin{array}{rcl}\text{Impurity}(n,N)& =& \frac{1}{n}+\frac{1}{n^2}+....+\frac{1}{n^N}\\ & =& \frac{1}{n}\frac{1-\frac{1}{n^N}}{1-\frac{1}{n^N}}\\ & =& \frac{1}{n^N}\frac{n^N-1}{n-1}.\\ & & \end{array}$

Therefore, for the two scenarios above, $n=2$and $n=3$, we'll look into the $\underset{N→∞}{\lim }$to determine what fraction of the impurity we can remove in an infinite number of steps.

First scenario: $n=2$

$\begin{array}{rcl}\underset{N→∞}{\lim }\text{Impurity}\left(2,N\right)& =& \underset{N→∞}{\lim }\frac{1}{2^N}\frac{2^N-1}{2-1}\\ \underset{N→∞}{\lim }\frac{2^N}{2^N}& =& 1\end{array}$

Second scenario: $n=3$

$\begin{array}{rcl}\underset{N→∞}{\lim }\text{Impurity}\left(3,N\right)& =& \underset{N→∞}{\lim }\frac{1}{3^N}\frac{3^N-1}{3-1}\\ \underset{N→∞}{\lim }\frac{3^N}{3^N}\frac{1}{2}& =& \frac{1}{2}\end{array}$

Thus, in the first scenario, we can eradicate all of the impurity; however, in the second scenario, with only a third removed at each stage, we can only hope to remove one half of the impurity.