If a limit exists it satisfies the relation: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:

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:

Base of Induction for n=1

Hypothesis of Induction for n=k

Step of Induction for n=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:

we notice

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:

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

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):

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

Since x represents we have:

because .

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

is true for all terms.

Base of Induction for n=1

Hypothesis of Induction for n=k

Step of Induction for n=k+1

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 where k is positive, is the positive root. (We dropped the negative root), this means that 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 where r is the root, to be correct

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

So the sequence is a null sequence because a_{n} always increases twoards therand never becomesr.

Since is a null sequence I can recall a theorem that says converges torif and only ifis 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!

which can easily be solved.

CB