Prove convergence and find the limit for the sequence $\displaystyle \{a_n\}$, where
$\displaystyle a_{n+1} = \frac{a_n^2+3}{2(a_n+1)}$ for $\displaystyle n\geq 1$, and $\displaystyle a_1=0.$
1) Prove that $\displaystyle a_n\geq 1\,\,\,\forall n$
2) Now prove that $\displaystyle a_n\leq 2\,\,\,\forall n$
3) Now prove that $\displaystyle \forall n\,,\,\,a_n\geq a_{n+1}$ . Hint: $\displaystyle \displaystyle{a_n\geq a_{n+1}=\frac{a_n^2+3}{2(a_n+1)}\Longleftrightarro w 2a_n^2+2a_n\geq a_n^2+3$ and etc. Use (1) above.
4) Now prove the seq. converges and using arithmetic of functions prove its limit is 1.
Tonio
The difference equation can be written in the form...
$\displaystyle \displaystyle \Delta_{n} = a_{n+1} - a_{n} = \frac{1}{2}\ \frac{3 - 2\ a_{n} - a^{2}_{n}}{1+a_{n}} = f(a_{n})$ (1)
The function f(x) is represented here...
There is only one 'attractive fixed point' in $\displaystyle x_{0}=1$ and any $\displaystyle a_{1}> -1$ will generate a sequence converging at $\displaystyle x_{0}$. If $\displaystyle -1< a_{1} < 1$ will be $\displaystyle a_{2}>1$ and the successive sequence will be monotonically decreasing...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$