Continuity of probability algebra help

Question

Suppose that $S = \{1, 2, 3,...\}$ is the set of all positive integers and that $P(\{s\})$ = $2−s$ for all $s \in S$. Compute P(A) where $A = \{2, 4, 6,...\}$ is the set of all even positive integers. Do this by using continuity of $P$

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Solution:

let $A_n = \{2,4,6,\cdots,2n\}$. Then by finite additivity, $$P(A_n) = P(2) + P(4) + \cdots + P(2n) = 2^{-2} + 2^{-4} + \cdots + 2^{-2n} = (1/4)\frac{[1-(1/4)^n]}{[1 - (1/4)]}$$

Hence, $P(A) = \lim_{n\to\infty} P(A_n) = \lim_{n\to\infty} (1/3)[1-(1/4)^n] = 1/3$

I don't understand the algebra. So they use the geometric series test $(1/4)\frac{[1-(1/4)^n]}{[1 - (1/4)]}$ and then it becomes $\lim_{n\to\infty} (1/3)[1-(1/4)^n] = 1/3$?

Mainly how to get the $(1/3)[1-(1/4)^n]$

Answer 1

Let $a=1/4$, then

$P(A_n)=a+a^2+...+a^n=a(1+a+...+a^{n-1})=$

$=a \frac{1-a^n}{1-a}=(1/4)\frac{[1-(1/4)^n]}{[1 - (1/4)]}=(1/4)\frac{[1-(1/4)^n]}{3/4}=(1/3)[1-(1/4)^n] \to 1/3$.

Answer 2

$(1/4)\frac{[1-(1/4)^n]}{[1 - (1/4)]}= (1/4)\frac{[1-(1/4)^n]}{[3/4]}=\frac {\frac 14}{\frac 34}{[1-(1/4)^n]}=\frac 13{[1-(1/4)^n]}$

That's where the $\frac 13$ comes from I guess. Not where babies are made. But hey. Fractions are cute :)