1. ## convergence

Prove convergence and find the limit for the sequence $\{a_n\}$, where
$a_{n+1} = \frac{a_n^2+3}{2(a_n+1)}$ for $n\geq 1$, and $a_1=0.$

2. Originally Posted by Sambit
Prove convergence and find the limit for the sequence $\{a_n\}$, where
$a_{n+1} = \frac{a_n^2+3}{2(a_n+1)}$ for $n\geq 1$, and $a_1=0.$
1) Prove that $a_n\geq 1\,\,\,\forall n$

2) Now prove that $a_n\leq 2\,\,\,\forall n$

3) Now prove that $\forall n\,,\,\,a_n\geq a_{n+1}$ . Hint: $\displaystyle{a_n\geq a_{n+1}=\frac{a_n^2+3}{2(a_n+1)}\Longleftrightarro w 2a_n^2+2a_n\geq a_n^2+3$ and etc. Use (1) above.

4) Now prove the seq. converges and using arithmetic of functions prove its limit is 1.

Tonio

3. Just practicing creating diagrams...

4. The difference equation can be written in the form...

$\displaystyle \Delta_{n} = a_{n+1} - a_{n} = \frac{1}{2}\ \frac{3 - 2\ a_{n} - a^{2}_{n}}{1+a_{n}} = f(a_{n})$ (1)

The function f(x) is represented here...

There is only one 'attractive fixed point' in $x_{0}=1$ and any $a_{1}> -1$ will generate a sequence converging at $x_{0}$. If $-1< a_{1} < 1$ will be $a_{2}>1$ and the successive sequence will be monotonically decreasing...

Kind regards

$\chi$ $\sigma$