The population of species at year $N$ is $P_{N}$. Assume that $P_{0}=1000$. $P$ is governed by the rule $P_{n+1} = 2P_{n}$ with probability $\frac{1}{2}$ and $P_{n+1} = 0$ with probability $\frac{1}{2}$. What is the expected time until the population dies out?
My idea: Let $M$ be the number of years when the population first becomes zero. Population can become zero at the 1st, 2nd,....nth,..., $\infty$ year. Then the expectation $E(M)$ is
$$E(M) = 1 (\frac{1}{2}) + 2(\frac{1}{2})^{2} + 3(\frac{1}{2})^{3} +..+ n(\frac{1}{2})^{n}+...$$
Summing this to infinity and taking the limit, using geometric series and using that $\frac{1}{1-x} = \sum x^{n}$ and $\frac{1}{(1-x)^{2}} = \sum n x^{n-1}$, we find that
$E(M) = 2$. So the expected time is $2$.
Is this correct?
As phrased, the population either grows or dies out at any point in time $n$, and after the population dies out it never returns. Thus we can imagine the process being a coin flip. On tails the population grows and on heads it dies forever. The question is if we flip a coin, how long will it take until our coin comes up heads (the population dies out)? This is a geometric random variable and the expected value of such a variable is $p^{-1} = (1/2)^{-1} = 2$, so the expected time is 2.