example of converging increasing sequence

Let $a$ be a positive real number and $q$ an integer greater than 1.  Set

$$x_1:=\sqrt[q]{a},$$

$$x_2:=\sqrt[q]{a+x_1}=\sqrt[q]{a+\sqrt[q]{a}},$$

$$x_3:=\sqrt[q]{a+x_2}=\sqrt[q]{a+\sqrt[q]{a+\sqrt[q]{a}}},$$

and generally

$x_n:=\sqrt[q]{a+x_{n-1}}.$ (1)

Since  $x_1>0$,  the two first above equations imply that  .  By induction on $n$ one can show that

The numbers $x_n$ are all below a finite bound $M$.  For demonstrating this, we write the inequality    in the form  , which implies   ,  i.e.

(2)

for all $n$.  We study the polynomial

$$f(x):=x^q-x-a=x(x^{q-1}-1)-1.$$

From its latter form we see that the function $f$ attains negative values when  $0\leqq x\leqq 1$  and that $f$ increases monotonically and boundlessly when $x$ increases from 1 to $\mathrm{\infty }$.  Because $f$ as a polynomial function is also continuous, we infer that the equation

$x^q-x-a=0$ (3)

has exactly one root (http://planetmath.org/Equation)   $x=M>1$,  and that $f$ is negative for    and positive for  $x>M$.  Thus we can conclude by (2) that    for all values of $n$.

The proven facts

settle, by the theorem of the parent entry (http://planetmath.org/NondecreasingSequenceWithUpperBound), that the sequence

$$x_1,x_2,x_3,\mathrm{\dots },x_n,\mathrm{\dots }$$

converges to a limit $x^{\prime }\leqq M$.

Taking limits of both sides of (1) we see that $x^{\prime }=\sqrt[q]{a+x^{\prime }}$,  i.e.  $x^{\prime q}-x^{\prime }-a=0$,  which means that  $x^{\prime }=M$,  in other words: the limit of the sequence is the only $M$ of the equation (3).

Title	example of converging increasing sequence
Canonical name	ExampleOfConvergingIncreasingSequence
Date of creation	2013-03-22 17:40:44
Last modified on	2013-03-22 17:40:44
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Example
Classification	msc 40-00
Related topic	NthRoot
Related topic	BolzanosTheorem