I have a probability below which denotes the chance of catching a fish.
$$P = \left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2 + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^4 + \dots $$
I can find a generalized form of $P$ by assuming the first term is $\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^0$ to come up with an infinite series:
$$P = \sum_{n=0}^\infty \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{2n}$$
I'm learning about series now and I know that if $|r| \geq 1$ the series is divergent. Here I have $|r| = \dfrac{3}{4}$ so there is a sum. I also have $a = \dfrac{1}{4}$. In the examples, the series have an exponent of $n$ but none have $2n$. Were it to be $n$ instead, I could find the sum as:
$$P = \frac{a}{1-r} = \frac{\frac{1}{4}}{1-\frac{3}{4}} = 1$$
But this is with $n$, not $2n$, where the answer I seek is $\dfrac{4}{7}$. How can I approach this number? How do I reconcile the $2n$ and is there a formula for this like there was when it was $n$? I can do partial sums and find an approximation but what is the sum as $n \to \infty$?
I've marked the question calculus as this is a unit in Stewart's Early Transcendentals calculus text.
Best Answer
$$\sum_{n=0}^\infty \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{2n}=\frac{1}{4}\sum_{n=0}^\infty\left(\frac{9}{16}\right)^{n}=\frac{1}{4} \frac{1}{1- \frac{9}{16}}=\frac{4}{7}.$$