$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$. Find its 10th term. Given this sequence, find its 10th term and its exact limit.
$$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$$
I've been stuck in this question forever. I can't find any relation between them. The answer for the 10th term is $\frac{5333}{3771}$
 A: Using the hint by @robjohn, the general solution of the recurrence is of the form
$$c_0r_0^n+c_1r_1^n$$ where the $r$ are the roots of the characteristic equation, $r^2-2r-1=0$. Unless the two roots are equal, the largest quickly dominates and the solution can be approximated as $c_0r_0^n$.
The initial conditions are
$$c_0+c_1=t_0,\\c_0r_0+c_1r_1=t_1,$$
from which you draw 
$$c_0(r_0-r_1)=t_1-t_0r_1.$$
Then the limit value of the fractions is
$$L=\frac{5-r_1}{3-2r_1}.$$
By completing the square, we have that $(r-1)^2=2$ and
$$L=\frac{4+\sqrt2}{1+2\sqrt2}=\sqrt2.$$
A: $$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$$
$$\frac{1}{2},\frac{5}{(2+1)},\frac{11}{(2+1+5)},\frac{27}{(2+1+5+11)},...$$
$$\frac{1}{2},\frac{(2+3)}{3},\frac{(3+8)}{8},\frac{(8+19)}{19},...$$
In this way I would define the series as follows:
$$\text{ for }  a_1=\frac{x_1}{y_1} = \frac{1}{2}$$
$$a_n := \frac{x_n}{y_n} = \frac{y_{(n-1)} + y_n}{x_{(n-1)}+y_{(n-1)}} $$
so 
$$a_n := \frac{x_n}{y_n} = \frac{y_{(n-1)} + y_n}{x_{(n-1)}+y_{(n-1)}} = \frac{y_{(n-1)} + \left(x_{(n-1)}+y_{(n-1)}\right)}{x_{(n-1)}+y_{(n-1)}} = \frac{2y_{(n-1)} + x_{(n-1)}}{x_{(n-1)}+y_{(n-1)}} =$$

$$ a_n = \left(1 + \frac{y_{(n-1)}}{y_{(n-1)}+x_{(n-1)}}\right)$$

the series until the 10th term:
$$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},\frac{65}{46}, \frac{157}{111}, \frac{379}{368}, \frac{915}{647}, \frac{2209}{1562},  \frac{5333}{3771}$$
A: Hint.
Notice the following pattern:
$$\begin{cases} a_{n+1} = b_n+b_{n+1} \\ b_{n+1} = a_n+b_n\end{cases}$$
if you denote by $$\frac{a_n}{b_n}$$ the $n$-th term of your sequence.
Now you can find a way to write your sequence in a non-recursive way in order to compute any term, and to find the limit.
