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Question about sequence

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
Hey!!! :)
I have a question.
It is given that $a>0 , x_{1}=x>0 \text{ and } x_{n+1}=\frac{1}{2}(x_{n}+\frac{a}{x_{n}})$ and I have to show that the sequence $(x_{n})$,at least from its second term,is decreasing and bounded from below from $\sqrt{a}$.Also,I have to find the limit $\lim_{n \to \infty}x_{n}$.

To show that the sequence is decreasing,I thought that I could take $\frac{x_{n+1}}{x_{n}}$ and show that it is smaller that $1$.I found that $\frac{x_{n+1}}{x_{n}}=\frac{1}{2}(1+\frac{a}{x_{n}^{2}})=\frac{1}{2}+\frac{a}{2x_{n}^{2}}$.
But do we know that it is $\frac{1}{2}+\frac{a}{2x_{n}^{2}}<1 \Rightarrow x_{n}>\sqrt{a}$ ?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
If the sequence is bounded below and decreasing, then the limit exists. So, assuming you've shown these two, in order to find the limit, just take the limit of both sides of the equation:
$$\lim_{n\to \infty}x_{n+1}= \lim_{n \to \infty} \frac12 \left(x_{n}+ \frac{a}{x_{n}} \right),$$
or if $\lim_{x\to \infty}x_{n}=L$, then
$$L= \frac12 (L+a/L).$$
Now just solve for $L$.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
If the sequence is bounded below and decreasing, then the limit exists. So, assuming you've shown these two, in order to find the limit, just take the limit of both sides of the equation:
$$\lim_{n\to \infty}x_{n+1}= \lim_{n \to \infty} \frac12 \left(x_{n}+ \frac{a}{x_{n}} \right),$$
or if $\lim_{x\to \infty}x_{n}=L$, then
$$L= \frac12 (L+a/L).$$
Now just solve for $L$.
I understand..Could I also say that because of the fact that the sequence is bounded below and decreasing,it converges to its inf??

Also,how can I show that it is decreasing? Can I use $\frac{x_{n+1}}{x_{n}} $ or do I have to do something else?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
I understand..Could I also say that because of the fact that the sequence is bounded below and decreasing,it converges to its inf??
Yes, any monotonically decreasing sequence that is bounded below has to converge.

Also,how can I show that it is decreasing? Can I use $\frac{x_{n+1}}{x_{n}} $ or do I have to do something else?
You could try to show that $x_{n+1}/x_{n}$ is less than $1$. That might work. What do you get when you try that?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
Yes, any monotonically decreasing sequence that is bounded below has to converge.
But we know that then the sequence converges to its infimum,or am I wrong? :confused:

You could try to show that $x_{n+1}/x_{n}$ is less than $1$. That might work. What do you get when you try that?
I found this: $\frac{1}{2}+\frac{a}{2x_{n}^{2}}$ ...But..Do we know that this is less than $1$ ?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
But we know that then the sequence converges to its infimum,or am I wrong? :confused:
Well, it depends on $x_{1}$. You probably could say that the sequence converges to its $ \lim \inf$.

I found this: $\frac{1}{2}+\frac{a}{2x_{n}^{2}}$ ...But..Do we know that this is less than $1$ ?
I think you'll have to take three cases: $x_{1}< \sqrt{a}$, $x_{1}= \sqrt{a}$, and $x_{1}> \sqrt{a}$. The middle case is fairly straight-forward: the sequence converges to $\sqrt{a}$ immediately, and never changes. Also, the case $x_{1}> \sqrt{a}$ is straight-forward: the sequence is monotonically decreasing. It's that pesky $x_{1}< \sqrt{a}$ case that's troubling... What do you make of it?
 

Krizalid

Active member
Feb 9, 2012
118
By induction. Let $n\ge1$. We have $\displaystyle{{x}_{2}}-\sqrt{a}=\frac{1}{2}\left( \frac{a}{{{x}_{1}}}+{{x}_{1}} \right)-\sqrt{a}=\frac{1}{2}\cdot \frac{{{({{x}_{1}}-\sqrt{a})}^{2}}}{{{x}_{1}}}>0,$ so this shows that $x_2>\sqrt a$. Suppose now that $x_n>\sqrt a,$ so $x_n>0$ and in the same fashion we have $\displaystyle{{x}_{n+1}}-\sqrt{a}=\frac{1}{2}\left( \frac{a}{{{x}_{1}}}+{{x}_{1}} \right)-\sqrt{a}=\frac{1}{2}\cdot \frac{{{({{x}_{n}}-\sqrt{a})}^{2}}}{{{x}_{n}}}>0,$ implying that $x_{n+1}>\sqrt a,$ so $x_n>\sqrt a$ for all $n\ge1.$ Let's now show that $x_{n+1}>x_n$ for all $n\ge1$. Since $0<\sqrt a<x_2$ we have $\displaystyle\frac{a}{{{x}_{2}}}-{{x}_{2}}=\frac{a-x_{2}^{2}}{{{x}_{2}}}<0\implies \frac{a}{{{x}_{2}}}+{{x}_{2}}<2{{x}_{2}}\implies {{x}_{3}}<{{x}_{2}}.$ In the same fashion for $0<\sqrt a<x_{n+1}$ we get $\displaystyle\frac{a}{{{x}_{n+1}}}-{{x}_{n+1}}=\frac{a-x_{n+1}^{2}}{{{x}_{n+1}}}<0\implies \frac{a}{{{x}_{n+1}}}+{{x}_{n+1}}<2{{x}_{n+1}},$ so ${{x}_{n+2}}<{{x}_{n+1}}.$

Finally $x_n>\sqrt a$ and $x_{n+1}<x_n,$ hence $x_n$ is a decreasing bounded below sequence, so it has a limit.
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
But we know that then the sequence converges to its infimum,or am I wrong? :confused:
Yes. A decreasing bounded below sequence converges to its infimum.