A population starts with one amoeba. In each generation, each amoeba divides in two with probability $\frac{1}{2}$, or dies, with probability $\frac{1}{2}$. Let $p_n$ be the probability that the population will survive $n$ generations.
$(a)$ What is $p_4$?
$(b)$ Find the limit $a = \lim_{n\to\infty} np_n$
$(c)$ Show that $p_n = \frac{a}{n} + \frac{b \space \log(n)}{n^2} + \mathcal{O} \left(\frac{1}{n^2}\right) $ as $n \to \infty$, and find $b$.
I have had a hard time trying to find a method to solve the problem so any help is much appreciated!
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ Each amoeba can generate $0$ amoebas or $2$ amoebas with the same probability $\pars{~{1 \over 2}~}$. Then, the probability $P_{n}$ that an amoeba generate $n$ amoebas in the next generation is given by $$ P_{n} = {1 \over 2}\pars{\delta_{n,0} + \delta_{n,2}} $$ The probability $P_{N \to N'}$ that $N$ amoebas generate $N'$ amoebas in the next generation is given by: \begin{align} P_{N \to N'} &= \left.\sum_{n_{1} = 0}^{\infty}P_{n_{1}}\ldots\sum_{n_{N} = 0}^{\infty}P_{n_{N}} \right\vert_{\sum_{i = 1}^{N}n_{i} = N'} \\[3mm]&= \sum_{n_{1} = 0}^{\infty}P_{n_{1}}\ldots\!\!\!\sum_{n_{N} = 0}^{\infty}P_{n_{N}} \int_{\verts{z} = 1}{1 \over z^{\pars{\sum_{i = 1}^{N}n_{i}\ -\ N'\ +\ 1}}} \,{\dd z \over 2\pi\ic} \\[3mm]&= \int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{1 \over z^{1 - N'}} \pars{\sum_{n = 0}^{\infty}P_{n}\,{1 \over z^{n}}}^{N} = \int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{1 \over z^{1 - N'}} \pars{{1 \over 2}\,{1 \over z^{0}} + {1 \over 2}\,{1 \over z^{2}}}^{N} \\[3mm]&= {1 \over 2^{N}}\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\, {1 \over z^{1 - N' + 2N}} \pars{1 + z^{2}}^{N} = {1 \over 2^{N}}\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\, {1 \over z^{1 - N' + 2N}} \sum_{n = 0}^{N}{N \choose n}z^{2n} \\[3mm]&= {1 \over 2^{N}}\sum_{n = 0}^{N}{N \choose n}\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\, {1 \over z^{1 - N' + 2N - 2n}} = {1 \over 2^{N}}\sum_{n = 0}^{N}{N \choose n} \delta_{2n,2N - N'} \end{align} The last sum is non zero whenever $$ 0 \leq 2N - N' \leq 2N\quad\mbox{and}\quad N'\ \mbox{even} $$ Then $$ P_{N \to N'} = \left\lbrace% \begin{array}{lcl} {1 \over 2^{N}}{N \choose \vphantom{\large A^{A}}{N' \over 2}} & \mbox{if} & \left\lbrace% \begin{array}{l} 0 \leq N' \leq 2N \\ N'\ \mbox{even} \end{array}\right. \\[2mm] 0 && \mbox{otherwise} \end{array}\right. $$ Once we know $P_{N \to N'}$ we can analyze several situations.