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

• Aug 10th 2011, 02:19 AM
Melsi
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:

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

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: $a_{n+1}> a_{n}$

Base of Induction for n=1

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

Hypothesis of Induction for n=k

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

Step of Induction for n=k+1

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

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: $a_{n}

we notice

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

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:

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

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

$a_{n}^{2}-a^{n}

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!!(Wink)

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

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

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

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

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

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

because $k>0$.

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
$a_{n}< \frac{\sqrt{4a+1}+1}{2}$
is true for all terms.

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

Hypothesis of Induction for n=k

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

Step of Induction for n=k+1

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

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 $a_{n}+k$ where k is positive, is the positive root. (We dropped the negative root), this means that $a_{n}$ 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!!!!(Surprised)

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

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

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

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

Since $(k_{n})$ is a null sequence I can recall a theorem that says $(a_{n})$ converges to r if and only if $(k_{n})$ 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!
• Aug 10th 2011, 05:00 AM
CaptainBlack
Re: Convergence and Limit Of Sequence, my Solution Too Big? Wonder If Is Good!
Quote:

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:

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

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: $a_{n+1}> a_{n}$

Base of Induction for n=1

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

Hypothesis of Induction for n=k

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

Step of Induction for n=k+1

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

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: $a_{n}

we notice

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

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:

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

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

$a_{n}^{2}-a^{n}

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!!(Wink)

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

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

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

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

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

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

because $k>0$.

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
$a_{n}< \frac{\sqrt{4a+1}+1}{2}$
is true for all terms.

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

Hypothesis of Induction for n=k

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

Step of Induction for n=k+1

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

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 $a_{n}+k$ where k is positive, is the positive root. (We dropped the negative root), this means that $a_{n}$ 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!!!!(Surprised)

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

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

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

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

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

$k=\sqrt{a+k}$

which can easily be solved.

CB
• Aug 10th 2011, 07:13 AM
Melsi
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:

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

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

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

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

$a_{n}^{2}-a_{n}

from which the value

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

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

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

is the limit.

My apologies..
• Aug 10th 2011, 07:51 AM
Melsi
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) $k=\sqrt{a+k}$

which can be analyzed like this:

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

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!

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

bye!(Clapping)
• Aug 10th 2011, 08:33 AM
chisigma
Re: Solved
The problem is to find [if it exists...] the limit of sequence satisfying the recurrence relation...

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

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

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

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

http://digilander.libero.it/luposabatini/MHF1200.bmp

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$
• Aug 10th 2011, 09:39 AM
Melsi
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
• Aug 11th 2011, 12:52 PM
chisigma
Re: Solved
Quote:

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...

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

... that can be written as...

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

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

http://digilander.libero.it/luposabatini/MHF126.bmp

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

Kind regards

$\chi$ $\sigma$
• Aug 12th 2011, 07:05 AM
Melsi
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:

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

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

(Yes)