Prove by Induction - Sequence The sequence $x_1, x_2, x_3, \ldots$ is such that $x_1 = 1 $ and
$$x_{n+1} \space = \frac{1+4x_n}{5 + 2x_n}$$
Prove by induction that $x_n > 0.5$ for all $n \ge 1$.
I have absolutely no clue how to go about this one. Can someone please explain.
Very sorry about the error, that's 0.5 and not 1.
 A: As others have commented, the problem as stated is (twice) incorrect, since $x_1 = 1$ and $x_2 = 5/7 < 1$, and it is therefore impossible to give an answer as is.
However, it is possible to make a complete study of the sequence $x_n$ in the following way: it is easy to see that if we define two sequences $y_n, z_n$ by
$$ y_1 = x_1, z_1 = 1, \quad y_{n+1} = 4 y_n + z_n, \quad z_{n+1} = 2 y_n + 5 z_n,$$
then we shall always have $x_n = y_n/z_n$. This is a linear recurrence with characteristic polynomial $r^2 - (4+5) r + (4\times 5-2\times 1) = r^2 - 9 r + 18 = (r-6) (r-3)$, which therefore has the characteristic roots $3$ and $6$.
So we write $y_n$ and $z_n$ as linear combinations of $3^n$ and $6^n$: we find
$$y_n = \frac{3^n+6^n)}{9},\quad z_n = \frac{-3^n+2\cdot 6^n}{9},$$
(manually check that the recurrence still holds!),
so that $$x_n = \frac{3^n+6^n}{-3^n+2\cdot 6^n} = \frac{1+2^n}{-1+2\cdot 2^n}$$.
(Again, manually check that this is actually $x_n$!).
This means that
 - the values taken by the sequence $(x_n)$ lie in the image by the function $t \mapsto (1+t)/(-1+2t)$ of the interval $[1,+\infty[$;
 - the sequence $x_n$ converges to $1/2$.
