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 analyticalor 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.

Thank you in advance!

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 therand never becomesr.

Since $\displaystyle (k_{n})$ is a null sequence I can recall a theorem that says $\displaystyle (a_{n})$ converges torif 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!