# sequence problems

• Dec 13th 2011, 12:55 AM
maoro
sequence problems
#1
Prove the sequence $\{x_n\}$ defined in the following is increasing and bounded.
$x_n=1+\frac{x_{n-1}}{1+x_{n-1}}$

I suppose to use MI to prove and let S(n) by the statement
$x_{n+1}\geq x_{n}$ and $1\leq x_n<2$
but i cant prove the sequence is increasing for the case n=k+1, am i starting with wrong statement ?

#2
Set $a_1=1$ ,and for $n\geq2$, $a_{n+1}=\frac{1}{2+a_n}$.Show that the sequence $\{a_n\}$ is convergent.

I am confusing how to let the statement.
• Dec 13th 2011, 01:27 AM
chisigma
Re: sequence problems
Quote:

Originally Posted by maoro
#1
Prove the sequence $\{x_n\}$ defined in the following is increasing and bounded...

$x_n=1+\frac{x_{n-1}}{1+x_{n-1}}$

The 'attack strategy' for this type of problems is illustrated in...

http://www.mathhelpforum.com/math-he...-i-188482.html

The recurrence relation defining the $x_{n}$ can be written as...

$\Delta_{n}= x_{n+1}-x_{n} = \frac{1+x_{n} - x_{n}^{2}} {1+x_{n}} = f(x_{n})$ (1)

The function f(x) has one 'attractive fixed point' in $x_{+}= \frac{1+\sqrt{5}}{2}$ and one 'repulsive fixed point' in $x_{-}= \frac{1-\sqrt{5}}{2}$ and because the inequality $|f(x)|< |x_{+}-x|$ all the initial values $x_{0}> x_{-}$ will produce a sequence converging to $x_{+}$. All the initial values $x_{0}< x_{-}$ however will produce a diverging sequence...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\chi$ $\sigma$

P.S. An important detail: all the initial values $x_{-} will produce an increasing sequence and all the initial values $x_{0}>x_{+}$ will produce a decreasing sequence...
• Dec 13th 2011, 05:08 AM
chisigma
Re: sequence problems
Quote:

Originally Posted by maoro
#2
Set $a_1=1$ ,and for $n\geq2$, $a_{n+1}=\frac{1}{2+a_n}$.Show that the sequence $\{a_n\}$ is convergent.

I am confusing how to let the statement.

The 'strategy' is the same illustrated in my previous post. The difference equation generating the $a_{n}$ can be written as...

$\Delta_{n}= a_{n+1}-a_{n}= \frac{1}{2+a_{n}}-a_{n}=f(a_{n})$ (1)

Here f(x) has two 'attractive fixed points' in $x_{-}=-1-\sqrt{2}$ and $x_{+}=-1+\sqrt{2}$. The attractive point $x_{-}$ however is 'pratically unarrivable' so that we indagate on $x_{+}$. For this 'attractive fixed point' the conditions of convergence are satisfied for any 'initial value' $a_{0}>-2$. For $-2 the convergence is 'oscillatory' and for $a_{0}>x_{+}$ the convergence will be 'monotonic'...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\chi$ $\sigma$
• Dec 13th 2011, 07:46 PM
maoro
Re: sequence problems
sorry, but are there easier alternatives to approach them ?