I am trying to solve an infinite ladder circuit. Let $\gamma_n$ denote the equivalent capacitance in the $n$th step. I arrived at the following recurrence relation linking $\gamma_{n-1}$ and $\gamma_n$. I am not sure how to get the equivalent capacitance from the recurrence relation so obtained.
$$\gamma_{n-1}=\frac{(\gamma _n+2^{n-1}\alpha)2^{n-2}\alpha}{\gamma_n+3\cdot2^{n-2}\alpha}$$
How to solve this recurrence relation? Since $n\to \infty$, $\gamma_{n-1}=\gamma_{n}$ but that would require putting in $n\to \infty$ into the entire expression, which seems to be indicating $\lim \gamma_n \to \infty$ and that might be true. So only taking the limit does not give the equivalent capacitance. Any hints are appreciated. Thanks
I think what you basically want to do is to solve the recurrence $x_{n+1}=\frac{x_n+1}{x_n+2},x_1=a$ where $a$ is given. If $a$ is rational, then $x_n$ will always be rational say of the form $\frac{p_n}{q_n}$ so $x_{n+1}=\frac{p_{n+1}}{q_{n+1}}=\frac{\frac{p_n}{q_n}+1}{\frac{p_n}{q_n}+2}=\frac{p_n+q_n}{p_n+2q_n}$. (Also note that if $p_n,q_n$ were coprime then the same is true of $p_n+q_n$ and $p_n+2q_n$.)
Thus the numerator and denominator satisfy the matrix recursion $\begin{bmatrix} p_{n+1} \\ q_{n+1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} p_n \\ q_n \end{bmatrix}$. This matrix recurrence can be solved explicitly by diagonalization, you wind up with
$$\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}^n=\begin{bmatrix} -\frac{1+\sqrt{5}}{2} & \frac{\sqrt{5}-1}{2} \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \left ( \frac{3-\sqrt{5}}{2} \right )^n & 0 \\ 0 & \left ( \frac{3+\sqrt{5}}{2} \right )^n \end{bmatrix} \begin{bmatrix} -\frac{\sqrt{5}}{5} & \frac{5-\sqrt{5}}{10} \\ \frac{\sqrt{5}}{5} & \frac{5+\sqrt{5}}{10} \end{bmatrix}.$$
and $\begin{bmatrix} p_{n+1} \\ q_{n+1} \end{bmatrix}$ is that thing times $\begin{bmatrix} p_1 \\ q_1 \end{bmatrix}$. You can see that for large $n$ everything concentrates onto the second eigenvector so you end up with $p_n/q_n \to \frac{\sqrt{5}-1}{2}$ regardless of the (rational) initial data. You can write an explicit closed form for the ratio from here if you want.