Infinite Series question 
The first one, the effective resistance is $2R$, then $5R/3$ then $13R/8$ etc....
My job is to find the pattern/equation so I can find the total resistance when $20$ resistors are connected. Of course I can do the long way by using this formula I created 
$R_n = (R_{n-1} \times R)/(R_{n-1} +R) +R$. Can you guys help me find an equation where I can just input the value of $n$ and simply get an answer. Thanks!
The second question is: What would I get if I had an infinite number of resistance.
Help is very much appreciated :)
 A: Let $R_n$ be the resistance of a circuit with $2n$ resistors, so $R_1 = 2R$.  The circuit with $2(n+1)$ resistors is obtained from the circuit with $2n$ resistors by placing one resistor in parallel and another one in series.  So
$$
R_{n+1} = R + \frac{1}{\frac{1}{R} + \frac{1}{R_n}} = R + \frac{R R_n}{R+R_n}.
$$
Substituting $R_n = \frac{p_n}{q_n}R$ gives
$$
R_{n+1} = R + \frac{R R_n}{R+R_n} = \frac{2p_n + q_n}{p_n + q_n}R
$$
and so we can take $p_{n+1} = 2p_n + q_n$ and $q_{n+1} = p_n+q_n$.  Placing these numbers in a sequence like
$$
q_1, p_1, q_2, p_2, q_3, p_3, \dotsc
$$
results in the Fibonacci sequence
$$1,2,3,5,8,13,\dotsc$$
for which explicit expressions can be found on the wp page (with a shifted index $n$).  In the limit the quotient $\frac{p_n}{q_n}$ of two consecutive Fibonacci numbers tends to the golden ratio $\varphi = \frac{1+\sqrt{5}}{2}$.
A: It is a fact of life that homographic transforms like
$$
u:x\mapsto R+\frac{Rx}{R+x},
$$
are conjugate to affine transforms. The first step to make use of this fact for the iterates of $u$ is to determine a fixed point of $u$, here
$$
x^*=R\frac{1+\sqrt5}2,
$$
and to rewrite everything in terms of $1/(x-x^*)$, here
$$
\frac1{u(x)-x^*}=b+\frac{a}{x-x^*},
$$
for some $(a,b)$ you will compute, hence
$$
\frac1{u(x)-x^*}-c=a\left(\frac1{x-x^*}-c\right),
$$
for some $c$ you will compute. In particular,
$$
\frac1{u^n(x)-x^*}-c=a^n\left(\frac1{x-x^*}-c\right).
$$
This yields an explicit formula for $R_n=u^{n-1}(2R)=u^n(\infty)$, namely
$$
R_n=x^*+\frac1{c(1-a^n)}.
$$
Since $x^*$ is known, note that one can skip everything above and identify $a$ and $c$ through the values of $R_1=2R$ and $R_2=\frac53R$, using
$$
c(1-a)=\frac1{R_1-x^*}=\frac1R\frac{3+\sqrt5}2,\qquad
c(1-a^2)=\frac1{R_2-x^*}=\frac3R\frac{7+3\sqrt5}2.
$$
Better still, use $a=1/u'(x^*)$ and the expression of $c(1-a)$ above, to get, for every $n\geqslant1$,
$$
R_n=(s-1)R+\frac1s\frac{s^2-1}{s^{2n}-1}R,\qquad s=\frac{3+\sqrt5}2.
$$
A: The second question is easier. Suppose that the resistance for infinite number of elementary parts is equal to $R_{\infty}$. Then it should not change if you add one more elementary part. This gives the equation
$$R_{\infty}=\frac{RR_{\infty}}{R+R_{\infty}}+R,$$
with the solution $\displaystyle R_{\infty}=\frac{1+\sqrt{5}}{2}R$.

Now concerning the first question, you should first check that the initial recurrence relation
$$R_{n+1}=\frac{RR_n}{R+R_n}+R,$$
being rewritten in terms of the variable
$$u_n=\frac{1}{R_n-R_{\infty}}+\frac{R+R_{\infty}}{R_{\infty}(2R+R_{\infty})},\tag{1}$$
is equivalent to
$$u_{n+1}=\frac{(R+R_{\infty})^2}{R^2}u_n.$$
(this is just a rephrasing of approach of Did in simpler terms). This has the obvious solution
$$u_{N}=\left(\frac{R+R_{\infty}}{R}\right)^{2(N-1)}u_1.\tag{2}$$
Now from (1), (2) and $R_1=2R$ one obtains a general formula for $R_N$.
