a>1, X1>sqr(a).
X(n+1) = (a+X(n)) / (1+X(n))
Prove that X1>X3>X5..............
and X2<X4<X6..............
Thank you
Perhaps a little clearer: $\displaystyle a>1\,,\,\,x_1>\sqrt{a}\,,\,\,x_{n+1}=\frac{a+x_n}{ 1+x_n}$.
a) Prove first that $\displaystyle \forall\,n\,\in\mathbb{N}\,,\,\,x_{2n-1}>\sqrt{a}>x_{2n}\,$ (Hint: For odd $\displaystyle n\,,\,\sqrt{a}<x_n=\frac{a+\frac{a+x_{n-2}}{1+x_{n-2}}}{1+\frac{a+x_{n-2}}{1+x_{n-2}}}$ $\displaystyle =\frac{2a+(a+1)x_{n-2}}{2x_{n-2}+(a+1)}\Longleftrightarrow$
$\displaystyle 4ax_{n-2}^2+4a(a+1)x_{n-2}+a(a+1)^2<4a^2+4a(a+1)x_{n-2}+(a+1)^2x_{n-2}^2 ...$ and etc. (induction, of course). For even n do something simmilar)
b) Next...induction, again and this time let us do the one with even indexes: we want to prove that
$\displaystyle \forall\,\mbox{ even }n\,\in\mathbb{N}\,,\,\,x_n<x_{n+2}\Longleftrighta rrow\,x_n<\frac{a+\frac{a+x_n}{1+x_n}}{1+\frac{a+x _n}{1+x_n}}$ $\displaystyle =\frac{2a+(a+1)x_n}{2x_n+(a+1)}\Longleftrightarrow$ $\displaystyle 2x_n^2+(a+1)x_n<2a+(a+1)x_n$.. and etc., and something simmilar to show $\displaystyle x_{2n-1}>x_{2n+1}$
In fact the above's enough to show that both sequences $\displaystyle \{x_{2n}\}\,,\,\{x_{2n-1}\}$ converge. Now you may want to show they both converge and to the same limit...
Tonio
Just an added note. You probably could show the claim by solving the difference equation exactly.
If you let $\displaystyle x_n = \sqrt{a} \cdot \frac{y_n - 1}{y_n + 1}$
then your difference equation becomes
$\displaystyle
y_{n+1} = - \frac{\sqrt{a}+1}{\sqrt{a}-1}\cdot y_n
$
which is linear with the solution
$\displaystyle y_n = c \left( -\, \frac{\sqrt{a}+1}{\sqrt{a}-1}\right)^n.$