Thread: Convergence and Limit Of Sequence, my Solution Too Big? Wonder If Is Good!

1. Convergence and Limit Of Sequence, my Solution Too Big? Wonder If Is Good!

Hello,

There is this problem about a sequence that I have already solved. I have two questions:

1.Why is my solution so big (kept short here). Due to the size I don't expect anything analytical or don't expect an answear at all.

2. A second concern though is described in blue at the end. Suspecting a small answear needed I would appreciate any (no matter how small) tip.

The general term of the sequence is:

$\displaystyle a_{n}=\sqrt{a+\sqrt{a+\sqrt{...\sqrt{+\sqrt{a}}}}}$

Proove convergence for and find the limit of the sequence.

Solution

For prooving convergence we suspect for the case we need
1. Increasing
2. Upper Bound Exists

Part one for Increasing will proove that:$\displaystyle a_{n+1}> a_{n}$

Base of Induction for n=1

$\displaystyle a_{1+1}=a_{2}=\sqrt{a+\sqrt{a}}> \sqrt{a} =a_{1}\Leftrightarrow a+\sqrt{a}> a$

Hypothesis of Induction for n=k

$\displaystyle a_{k}=\sqrt{a+a_{k-1}}> a_{k-1}=\sqrt{a+a_{k-2}}$

Step of Induction for n=k+1

$\displaystyle a_{k+1}>a_{k}\Leftrightarrow \sqrt{a+\sqrt{a+a_{k-1}}}>\sqrt{a+\sqrt{a+a_{k-2}}}\Leftrightarrow a+\sqrt{a+a_{k-1}}>a+\sqrt{a+a_{k-2}}\Leftrightarrow a_{k}>a_{k-1}$

We started from the hypothesis of the step and found the hypothesis of the induction which is true by hypothesis. This makes the step true. This makes the claim true for every term of the sequence. So the sequence is originaly increasing.

Part two for bounded will proove that:$\displaystyle a_{n}<b$

we notice

$\displaystyle a_{3}^{2}=a+a^{2} \Leftrightarrow a_{3}^{2}-a^{2}=a$

this is a good help because we have some terms bounded by a constant, although it is an equality (we would like inequality) is good for now.

(1) We hope this equality is true for all terms, if yes the it is:

$\displaystyle a_{n}^{2}-a^{n-1}=a$

Now I need inequality so I can speak about an upper boundary:

$\displaystyle a_{n}^{2}-a^{n}<a$

it is true because the sequence is increasing. Good for now but I need to replace-find a constant expression for the n-th term because the upper bound now does not bounds the sequence itself but an expression of its terms.

Fortunately I have a polynomial of second degree and I have easy tools for finding constant root!!

I have the right to turn the inequality to equality (call me 2-nd degree equation):

$\displaystyle (a_{n}+k)^{2}-(a^{n}+k)=a$

Now I can solve the equation, leaving the negative root out because the terms are non negative:

$\displaystyle x^{2}-x=a\Leftrightarrow 4x^{2}-4x=4a\Leftrightarrow 4x^{2}-4x+1=4a+1 \Leftrightarrow 2x^{2}-2(2x)(1)+1^{2}=4a+1\Leftrightarrow (2x-1)^{2}=4a+1\Leftrightarrow 2x-1=\sqrt{4a+1}\Leftrightarrow 2x=\sqrt{4a+1}+1 \Leftrightarrow x=\frac{\sqrt{4a+1}+1}{2}$

Since x represents $\displaystyle a_{n}+k$ we have:

$\displaystyle a_{n}+k=\frac{\sqrt{4a+1}+1}{2}\Leftrightarrow a_{n}< \frac{\sqrt{4a+1}+1}{2}$

because $\displaystyle k>0$.

Now there is something else, at the point (1) above I made the hypothesis that my claim will be true for every term, this means I have to proove that
$\displaystyle a_{n}< \frac{\sqrt{4a+1}+1}{2}$
is true for all terms.

Base of Induction for n=1
$\displaystyle a_{1}<\frac{1+\sqrt{4a+1}}{2}\Leftrightarrow ...etc...\Leftrightarrow \sqrt{a}< \sqrt{a+\frac{1}{4}}+\frac{1}{2}$

Hypothesis of Induction for n=k

$\displaystyle a_{k}<\frac{1+\sqrt{4a+1}}{2}$

Step of Induction for n=k+1

$\displaystyle a_{k+1}<\frac{1+\sqrt{4a+1}}{2}\Leftrightarrow ..etc..\Leftrightarrow a+a_{k}< \frac{1^{2}+2\sqrt{4a+1}+\sqrt{4a+1}^2}{4 }\Leftrightarrow ...etc...\Leftrightarrow a_{k}<\frac{1+\sqrt{4a+1}}{2}$

So we prooved that every term of the sequence is bounded.

This means that the sequence does converge!

Part three calculating the limit

We found that $\displaystyle a_{n}+k$ where k is positive, is the positive root. (We dropped the negative root), this means that $\displaystyle a_{n}$ is found always on the left of the most right root because k is positive. Also because terms are positive they cannot be found elsewhere (for example on the left of the left most root, the negative root). This means that the terms will be found always between the two roots!!!!

This conclusion is a big help, because the equality $\displaystyle a_{n}+k=r$ where r is the root, to be correct $\displaystyle r=\frac{1+\sqrt{4a+1}}{2}$

I was saying this equality can help me build a null sequence that converges to zero like this:

$\displaystyle a_{n}+k =\frac{1+\sqrt{4a+1}}{2} \Leftrightarrow a_{1}+k_{1}=r, a_{2}+k_{2}=r,...etc..., a_{n}+k_{n}=r \Leftrightarrow r-a_{1}=k_{1},r-a_{2}=k_{2},...etc...,r-a_{n}=k_{n}$

So the sequence $\displaystyle (k_{n})$ is a null sequence because a_{n} always increases twoards the r and never becomes r.

Since $\displaystyle (k_{n})$ is a null sequence I can recall a theorem that says $\displaystyle (a_{n})$ converges to r if and only if $\displaystyle (k_{n})$ is a null sequence.

My problem now is that: what happens in case that there is a number k/2 where k is the same positive k refered here, that also claims to be a limit.

I suspect that this probably is not true, but cannot tell why! I tried induction and I think it worked, showed that r is the real limit and not some k/2. But is there any other way out of induction because I have already used induction too much and it seems not that good to me!

2. Re: Convergence and Limit Of Sequence, my Solution Too Big? Wonder If Is Good!

Originally Posted by Melsi
Hello,

There is this problem about a sequence that I have already solved. I have two questions:

1.Why is my solution so big (kept short here). Due to the size I don't expect anything analytical or don't expect an answear at all.

2. A second concern though is described in blue at the end. Suspecting a small answear needed I would appreciate any (no matter how small) tip.

The general term of the sequence is:

$\displaystyle a_{n}=\sqrt{a+\sqrt{a+\sqrt{...\sqrt{+\sqrt{a}}}}}$

Proove convergence for and find the limit of the sequence.

Solution

For prooving convergence we suspect for the case we need
1. Increasing
2. Upper Bound Exists

Part one for Increasing will proove that:$\displaystyle a_{n+1}> a_{n}$

Base of Induction for n=1

$\displaystyle a_{1+1}=a_{2}=\sqrt{a+\sqrt{a}}> \sqrt{a} =a_{1}\Leftrightarrow a+\sqrt{a}> a$

Hypothesis of Induction for n=k

$\displaystyle a_{k}=\sqrt{a+a_{k-1}}> a_{k-1}=\sqrt{a+a_{k-2}}$

Step of Induction for n=k+1

$\displaystyle a_{k+1}>a_{k}\Leftrightarrow \sqrt{a+\sqrt{a+a_{k-1}}}>\sqrt{a+\sqrt{a+a_{k-2}}}\Leftrightarrow a+\sqrt{a+a_{k-1}}>a+\sqrt{a+a_{k-2}}\Leftrightarrow a_{k}>a_{k-1}$

We started from the hypothesis of the step and found the hypothesis of the induction which is true by hypothesis. This makes the step true. This makes the claim true for every term of the sequence. So the sequence is originaly increasing.

Part two for bounded will proove that:$\displaystyle a_{n}<b$

we notice

$\displaystyle a_{3}^{2}=a+a^{2} \Leftrightarrow a_{3}^{2}-a^{2}=a$

this is a good help because we have some terms bounded by a constant, although it is an equality (we would like inequality) is good for now.

(1) We hope this equality is true for all terms, if yes the it is:

$\displaystyle a_{n}^{2}-a^{n-1}=a$

Now I need inequality so I can speak about an upper boundary:

$\displaystyle a_{n}^{2}-a^{n}<a$

it is true because the sequence is increasing. Good for now but I need to replace-find a constant expression for the n-th term because the upper bound now does not bounds the sequence itself but an expression of its terms.

Fortunately I have a polynomial of second degree and I have easy tools for finding constant root!!

I have the right to turn the inequality to equality (call me 2-nd degree equation):

$\displaystyle (a_{n}+k)^{2}-(a^{n}+k)=a$

Now I can solve the equation, leaving the negative root out because the terms are non negative:

$\displaystyle x^{2}-x=a\Leftrightarrow 4x^{2}-4x=4a\Leftrightarrow 4x^{2}-4x+1=4a+1 \Leftrightarrow 2x^{2}-2(2x)(1)+1^{2}=4a+1\Leftrightarrow (2x-1)^{2}=4a+1\Leftrightarrow 2x-1=\sqrt{4a+1}\Leftrightarrow 2x=\sqrt{4a+1}+1 \Leftrightarrow x=\frac{\sqrt{4a+1}+1}{2}$

Since x represents $\displaystyle a_{n}+k$ we have:

$\displaystyle a_{n}+k=\frac{\sqrt{4a+1}+1}{2}\Leftrightarrow a_{n}< \frac{\sqrt{4a+1}+1}{2}$

because $\displaystyle k>0$.

Now there is something else, at the point (1) above I made the hypothesis that my claim will be true for every term, this means I have to proove that
$\displaystyle a_{n}< \frac{\sqrt{4a+1}+1}{2}$
is true for all terms.

Base of Induction for n=1
$\displaystyle a_{1}<\frac{1+\sqrt{4a+1}}{2}\Leftrightarrow ...etc...\Leftrightarrow \sqrt{a}< \sqrt{a+\frac{1}{4}}+\frac{1}{2}$

Hypothesis of Induction for n=k

$\displaystyle a_{k}<\frac{1+\sqrt{4a+1}}{2}$

Step of Induction for n=k+1

$\displaystyle a_{k+1}<\frac{1+\sqrt{4a+1}}{2}\Leftrightarrow ..etc..\Leftrightarrow a+a_{k}< \frac{1^{2}+2\sqrt{4a+1}+\sqrt{4a+1}^2}{4 }\Leftrightarrow ...etc...\Leftrightarrow a_{k}<\frac{1+\sqrt{4a+1}}{2}$

So we prooved that every term of the sequence is bounded.

This means that the sequence does converge!

Part three calculating the limit

We found that $\displaystyle a_{n}+k$ where k is positive, is the positive root. (We dropped the negative root), this means that $\displaystyle a_{n}$ is found always on the left of the most right root because k is positive. Also because terms are positive they cannot be found elsewhere (for example on the left of the left most root, the negative root). This means that the terms will be found always between the two roots!!!!

This conclusion is a big help, because the equality $\displaystyle a_{n}+k=r$ where r is the root, to be correct $\displaystyle r=\frac{1+\sqrt{4a+1}}{2}$

I was saying this equality can help me build a null sequence that converges to zero like this:

$\displaystyle a_{n}+k =\frac{1+\sqrt{4a+1}}{2} \Leftrightarrow a_{1}+k_{1}=r, a_{2}+k_{2}=r,...etc..., a_{n}+k_{n}=r \Leftrightarrow r-a_{1}=k_{1},r-a_{2}=k_{2},...etc...,r-a_{n}=k_{n}$

So the sequence $\displaystyle (k_{n})$ is a null sequence because a_{n} always increases twoards the r and never becomes r.

Since $\displaystyle (k_{n})$ is a null sequence I can recall a theorem that says $\displaystyle (a_{n})$ converges to r if and only if $\displaystyle (k_{n})$ is a null sequence.

My problem now is that: what happens in case that there is a number k/2 where k is the same positive k refered here, that also claims to be a limit.

I suspect that this probably is not true, but cannot tell why! I tried induction and I think it worked, showed that r is the real limit and not some k/2. But is there any other way out of induction because I have already used induction too much and it seems not that good to me!
If a limit exists it satisfies the relation:

$\displaystyle k=\sqrt{a+k}$

which can easily be solved.

CB

3. It's getting somewhere!

Thank you Captain!

Well I am looking on it, but sorry I am a bit stuck I cannot put it all together,

you mean:

$\displaystyle \lim_{n \to \infty}a_{n}=k=\lim_{n \to \infty}a_{n+1}=\lim_{n \to \infty}\sqrt{a+a_{n}} = \sqrt{a+\lim_{n \to \infty}a_{n}} =\sqrt{a+k}\Leftrightarrow k=\sqrt{a+k}$

I didn't notice this but I understand it well now. I also understand that because of

$\displaystyle a_{n}^{2}-a_{n-1}=a$

where a can become as small as possible then the other relation

$\displaystyle a_{n}^{2}-a_{n}<a$

from which the value

$\displaystyle \frac{1+\sqrt{4a+1}}{2}$

results should force this value to be a limit.

Sorry! I am afraid I am still missing something, is this what you meant? I don't feel sure which part exactly assures me that

$\displaystyle \frac{1+\sqrt{4a+1}}{2}$

is the limit.

My apologies..

4. Solved

Hello again!

I think your tip brought a solution in my head suddenly (hope so..)!

It is true that if k is the limit then it must respect this:

(1) $\displaystyle k=\sqrt{a+k}$

which can be analyzed like this:

$\displaystyle k=\sqrt{a+k}\Leftrightarrow k^{2}=a+k\Leftrightarrow k^{2}-k=a\Leftrightarrow k^{2}-k-a=0$

and this is familiar because we already have dealt with this equation before!

Now there are these observations:

• the relation (1) is true only for the limit k
• the solution of the equation is true only for the root \frac{1+\sqrt{4a+1}}{2}
• we showed the equation resulted from the relation (1)

these observations mean that because k respects the relation (1) it is possible to be the root of the equation which also is unique, this means k is the root of the equation.

In other words the root of the equation is k, the limit!

$\displaystyle root=k\Leftrightarrow k=\frac{1+\sqrt{4a+1}}{2}$

bye!

5. Re: Solved

The problem is to find [if it exists...] the limit of sequence satisfying the recurrence relation...

$\displaystyle a_{n+1}= \sqrt{a + a_{n}}$ (1)

... with the 'initial condition' [that's important...] $\displaystyle a_{0}=a > 0$. The (1) can be written as...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= \sqrt{a + a_{n}} -a_{n}= f(a_{n})$ (2)

The $\displaystyle f(x)$ for the particular case $\displaystyle a=1$ is represented here...

We define 'attractive fixed point' a zero of $\displaystyle f(x)$ where $\displaystyle f(x)$ crosses the x axis with negative slope. In this case the only 'attractive fixed point' of $\displaystyle f(x)$ is at...

$\displaystyle x_{0}= \frac{1 + \sqrt{1+4\ a}}{2}$ (3)

Now the sequence tends to $\displaystyle x_{0}$ when $\displaystyle n$ tends to infinity if and only if...

a) $\displaystyle x_{0}$ is 'attractive fixed point'...

b) for the initial value $\displaystyle a_{0}$ is $\displaystyle |f(a_{0})|< |2\ (x_{0}-a_{0})|$ [the 'red line' in the figure...]

In our case a) and b) are satisfied...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Re: Solved

Never heard of attractive fixed points before, I had to check wikipedia for it (it had a good example, continously pressing the cos button on the calculator acts as a fixed point), so this is what is deeply hidden behind it, good to know, geometric view helps.

Thank you a lot!

I see.. χ σ... Χι Σίγμα

7. Re: Solved

Originally Posted by Melsi
Never heard of attractive fixed points before, I had to check wikipedia for it (it had a good example, continously pressing the cos button on the calculator acts as a fixed point), so this is what is deeply hidden behind it, good to know, geometric view helps...
The example reported in wikipedia is very instructive. Here the recursive relation is...

$\displaystyle a_{n+1}=\cos a_{n}$ (1)

... that can be written as...

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

The function $\displaystyle f(x)= \cos x - x$ is represented here...

We have only one 'attractive fixed point' in $\displaystyle x_{0}= .739085133215...$ and the condition $\displaystyle |f(x)| < |2\ (x_{0}-x)|$ [the 'red line'...] guarantees in any case the convergence at $\displaystyle x_{0}$. However the fact that is not verified the condition $\displaystyle |f(x)|<|x_{0}-x|$ [the 'blue line'...]implies that the convergence is not 'monotonic' but 'oscillating' as everibody can verify...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. Re: Solved

Very good graph and explanatory... this is so much related to the sequence I was working on.Again there is this (how can I say), the function encapsulating itself, which results to:

$\displaystyle f(x)=cos(a_{n})-a_{n}$ or $\displaystyle f(x)=\sqrt{a_{n}}-a_{n}$

I find the red line helps understanding a lot cause $\displaystyle x_{0}-x$ can become small enough to allow the red line cross the attractive point.