1. ## sequence

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

2. Originally Posted by felixmcgrady
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: $a>1\,,\,\,x_1>\sqrt{a}\,,\,\,x_{n+1}=\frac{a+x_n}{ 1+x_n}$.

a) Prove first that $\forall\,n\,\in\mathbb{N}\,,\,\,x_{2n-1}>\sqrt{a}>x_{2n}\,$ (Hint: For odd $n\,,\,\sqrt{a} $=\frac{2a+(a+1)x_{n-2}}{2x_{n-2}+(a+1)}\Longleftrightarrow$

$4ax_{n-2}^2+4a(a+1)x_{n-2}+a(a+1)^2<4a^2+4a(a+1)x_{n-2}+(a+1)^2x_{n-2}^2 ...$ 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

$\forall\,\mbox{ even }n\,\in\mathbb{N}\,,\,\,x_n $=\frac{2a+(a+1)x_n}{2x_n+(a+1)}\Longleftrightarrow$ $2x_n^2+(a+1)x_n<2a+(a+1)x_n$.. and etc., and something simmilar to show $x_{2n-1}>x_{2n+1}$

In fact the above's enough to show that both sequences $\{x_{2n}\}\,,\,\{x_{2n-1}\}$ converge. Now you may want to show they both converge and to the same limit...

Tonio

3. Originally Posted by felixmcgrady
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
Just an added note. You probably could show the claim by solving the difference equation exactly.

If you let $x_n = \sqrt{a} \cdot \frac{y_n - 1}{y_n + 1}$

then your difference equation becomes

$
y_{n+1} = - \frac{\sqrt{a}+1}{\sqrt{a}-1}\cdot y_n
$

which is linear with the solution

$y_n = c \left( -\, \frac{\sqrt{a}+1}{\sqrt{a}-1}\right)^n.$