# Real analysis problem with set convergence

1. Mar 10, 2009

### geoman

Define the sequence A where Asub1=a and Asubn=sqrt(a+Asub(n-1)) for n greater than or equal to 2
I need to determine what positive choices of a will make the sequence converge and to what limit. I also need to prove it.

I plugged some values into the sequence and found that it seems to converge for all positive real numbers but I don't know how find out what any given choice of a will converge to

2. Mar 11, 2009

### HallsofIvy

Staff Emeritus
If that sequence converges, then taking the limit of both sides of $A_n= \sqrt{a+ A_{n-1}}$, we have, because square root is continuous, $\lim_{n\rightarrow\infty}A_n= \sqrt{a+ \lim_{n\rightarrow\infty}A_{n-1}}$. Of course, those two limits are of the same sequence so, calling that limit L, $L= \sqrt{a+ L}$. Squaring both sides, $L^2= a+ L$ or $L^2- L- a= 0$. By the quadratic formula, $L= (1\pm\sqrt{1+ 4a})/2$. Which of the two roots (assuming 1+4a> 0) is the limit will depend on a.

(What does this have to do with "set convergence"? I see no sets here.)

3. Mar 11, 2009

### poutsos.A

Is the sequence you are considering the following??

1) $$x_{1} =a$$ .......

2) $$x_{n} = \sqrt(a + x_{n-1})$$.....for $$n\geq 2$$

4. Mar 11, 2009

### geoman

yes that is the sequence. And sorry I did not meant to say set convergence. I meant convergence of sequences.

5. Mar 16, 2009

### poutsos.A

geoman

This is a complete study of the convergence of the sequence:

PART A
Assume the sequence converges,then $$\lim _{ n\rightarrow\infty}\ x_n=x$$ and thus;

$$x=\sqrt{a + x}$$

And the solutions of the equation are:

$$y_{1} = \frac{1+\sqrt{1+4a}}{2}$$ ........................................................1

$$y_{2}=\frac{1-\sqrt{1+4a}}{2}$$..............................................................2

Now we have the following cases:

a) For a< -1/4 one of the subsequences of the sequence might have a limit on complex Nos .This is a point for research.

b) For $$-\frac{1}{4}\leq a<0$$ ,the sequence has no limit

c) For a=0 the sequence obviously has 0 as a limit ,although the equation of the sequence give us two limits 1 and 0.

d)For a>0 the equation has as its limit the positive root (1) for the following reasons:

We can prove by induction that $$x_{n}>0 \forall n\in N$$.......................................................................................................3

Also a>0 ===> 4a+1>1 =====> root (2) is negative.Hence if the above sequence converges then its limit is the positive root (1)

So we have proved:

(the sequence{$$x_{n}$$} converges)=====>(converges to root (1))

6. Mar 16, 2009

### poutsos.A

PART B

Now we must prove that the sequence converges

For that we must form two differences:

The difference $$x_{n+1}-x_{n}$$ and the difference $$x_{n+1}-y_{1}$$

Thus:

$$x_{n+1}-x_{n}=\sqrt{a+x_{n}}-x_{n} =\frac{a+x_{n}-x_{n}^2}{\sqrt{a+x_{n}}+x_{n}} =-\frac{(x_{n}-y_{1})(x_{n}-y_{2})}{\sqrt{a+x_{n}}+x_{n}}$$....................................................................................................4

Note $$(x_{n}-y_{2})>0\forall n\in N$$ since root (2) is negative .

That difference will show us whether the sequence increases or decreases and that will depend on the factor $$x_{n}-y_{1}$$ since all the other factors in (4) are positive.

For that we form the difference:

$$x_{n+1}-y_{1}= \sqrt{a+x_{n}}-y_{1} =\frac{a+x_{n}-y_{1}^2}{\sqrt{a+x_{n}}+y_{1}}$$,and if we substitute the value of $$y_{1}$$ we get:

$$x_{n+1}-y_{1}=\frac{2x_{n}-1-\sqrt{1+4a}}{2(\sqrt{a+x_{n}}+y_{1})}$$.........................................................................................................5

That difference will help us prove by induction whether $$x_{n}-y_{1}$$ is positive or negative and hence establish thru (4) whether the sequence is increasing or decreasing.

Indeed:

For $$x_{1}=a>y_{1}$$ we have if $$x_{n}>y_{1}$$ then $$x_{n}-y_{1}= 2x_{n}-1-\sqrt{1+4a}>0$$ and using (5) $$x_{n+1}-y_{1}>0$$,hence

$$x_{n}-y{1}>0 \forall n\in N$$..............................................................6

And

For $$x_{1}=a< y_{1}$$ we have if $$x_{n}< y_{1}$$ then $$x_{n}-y_{1}= 2x_{n}-1-\sqrt{1+4a}<0$$ and using (5) $$x_{n+1}-y_{1}<0$$,hence

$$x_{n}-y{1}<0 \forall n\in N$$..................................................................7

So in either case by substituting (6) or (7) into (4) we see that the sequence is increasing or decreasing and also either bounded from above or below ,since by (6) or (7):

$$x_{n}\leq y_{1}$$ or $$x_{n}\geq y_{1}$$ for all nεN

THUS the sequence converges and since in part A WE proved if it converges it converges to root (1), we conclude:

For a>0 the sequence converges to the positive root (1)

7. Mar 16, 2009

### poutsos.A

In part A OF the convergence of the sequence the equation $$x=\sqrt{a+x}$$ give us two solutions,and for a>0 we have two solutions ,one -ve and one +ve.

So one may wonder why the sequence converges to the +ve root ,due to the fact that,$$x_{n}>0 \forall n\in N$$ and not to the -ve root.The following proof shows why:

Suppose the sequence tends to the -ve root $$y_{2}$$.

From the definition of the limit we have :

For all ε>0 ,there exists a kεN SUCH that for all $$n\geq k$$ ,then $$|x_{n}-y_{2}|$$<ε

Ιn the above definition put ε= $$-y_{2}$$........... n=k,then $$|x_{k}-y_{2}|< -y_{2}$$ which implies $$x_{k}<0$$

But $$x_{n}>0 \forall n\in N$$ ,and for n=k we get :$$x_{k}>0$$.

therefor we end with the contradiction $$x_{k}>0$$ and $$x_{k}<0$$.

THUS we deduce that the sequence converges to the +ve root