Recurrence relation with reciprocal in a circuit The motivation for this recurrence relation is to find the total resistance in this circuit: 
Assuming that the capacitor has no resistance, with only one loop of the circuit, (let us suppose) the total resistance amounts to $R_{T_1} = R_o + R_m + R_i$. Then adding on the other loops successively one gets the recurrence relation:
$$R_{T_n} = R_o + R_i + \frac1{R_m^{-1} + R_{T_{n-1}}^{-1}}$$
Computationally I have deduced that the sequence $R_{T_n}$ converges to a certain value. How would I derive a closed form for the sequence from this recurrence relation, and therefore find the limiting value of the total circuit resistance (for an infinitely or sufficiently long neuron, not that that exists)?

Let $A = R_m$ and $\displaystyle\;\sigma = \frac{R_i + 2R_m + R_o}{R_m}$,  and writing $R_{T_n}$ as $\displaystyle\;A \frac{p_n}{q_n}$ for some $p_n, q_n$ whose overall scale doesn't matter, we have
$$R_{T_{n+1}} = A \frac{p_{n+1}}{q_{n+1}} = \underbrace{A(\sigma-2)}_{R_i+R_o} + \frac{1}{\underbrace{A^{-1}}_{R_m^{-1}} + \underbrace{A^{-1}\frac{q_n}{p_n}}_{R_{T_n}^{-1}}}
\quad\iff\quad
\frac{p_{n+1}}{q_{n+1}}
= \frac{(\sigma-1) p_n + (\sigma-2) q_n}{p_n + q_n}$$
We can transform this to a matrix form:
$$\begin{bmatrix}p_{n+1}\\q_{n+1}\end{bmatrix} \propto
\begin{bmatrix}\sigma-1 & \sigma-2\\1 & 1\end{bmatrix}
\begin{bmatrix}p_{n}\\q_{n}\end{bmatrix}
$$
It is clear the initial condition $R_{T_1} = R_i+R_m+R_o$ is equivalent to
$$
\frac{p_1}{q_1} = \sigma-1
\quad\iff\quad
\begin{bmatrix}p_{1}\\q_{1}\end{bmatrix}\propto
\begin{bmatrix}\sigma-1\\ 1\end{bmatrix}
= \begin{bmatrix}\sigma-1 & \sigma-2\\1 & 1\end{bmatrix}
\begin{bmatrix}1\\ 0\end{bmatrix}
$$
This implies for any $n > 0$,
$$\begin{bmatrix}p_{n}\\q_{n}\end{bmatrix} 
\propto 
\begin{bmatrix}\sigma-1 & \sigma-2\\1 & 1\end{bmatrix}^n 
\begin{bmatrix}1\\ 0\end{bmatrix}$$
The matrix $\begin{bmatrix}\sigma-1 & \sigma-2\\1 & 1\end{bmatrix}$ has
eigenvalues $\lambda_{\pm}$ and eigenvectors $u_{\pm}$ given by:
$$\lambda_{\pm} = \frac12\left(\sigma \pm \sqrt{\sigma^2-4}\right),
\quad
u_{\pm} = \begin{bmatrix}\lambda_{\pm} -1\\1\end{bmatrix}$$
Notice 
$$\begin{bmatrix}1\\ 0\end{bmatrix} \propto u_{+} - u_{-}
\quad\implies\quad
\begin{bmatrix}p_{n}\\q_{n}\end{bmatrix} \propto \lambda_{+}^n u_{+} - \lambda_{-}^n u_{-}$$
We find
$$R_{T_n} = R_m \frac{p_{n}}{q_{n}} = R_m
\frac{
\lambda_{+}^n(\lambda_{+} - 1) - 
\lambda_{-}^n(\lambda_{-} - 1)
}{
\lambda_{+}^n - \lambda_{-}^n
}
= R_m \left(\frac{
\lambda_{+}^{n+1} - \lambda_{-}^{n+1}
}{
\lambda_{+}^{n} - \lambda_{-}^n
} - 1\right)
$$
Since $|\lambda_{+}| > 1 > |\lambda_{-}|$, the RHS will be dominated by the $\lambda_{+}$ term when $n$ is large. This implies 
$$\begin{align}
\lim_{n\to\infty} R_{T_n} = R_m\lim_{n\to\infty} \frac{p_n}{q_n}
= R_m(\lambda_{+} - 1 )
&= \frac12\left( R_i + R_o + \sqrt{(R_i+2R_m+R_o)^2 - 4R_m^2}\right)\\
&= \frac12\left(R_i+R_o + \sqrt{(R_i+R_o)(R_i+R_o+4R_m)}\right)
\end{align}
$$
The same result you will get if you solve the quadratic equation mentioned in another answer.
