Proving Increasing Sequence or Showing Convergence of {a_n}

[mynameisfunk]

Homework Statement

Suppose that {$a_n$} is a bounded sequence of real numbers such that, for all $n$, $a_n \leq \frac{a_{n-1}+a_{n+1}}{2}$. Show that $b_n=a_{n+1}-a_n$ is an increasing sequence. Otherwise show that {$a_n$} converges.

Homework Equations

The Attempt at a Solution

I do not see it at all..
It seems that since $2a_n \leq a_{n-1}+a_{n+1}$ that the sequence could be either increasing OR decreasing. i could fit the set of integers, either increasing or decreasing and the inequality holds and $b_n$ is stagnate. BUT, since {$a_n$} is bounded, it must be convergent?? i couldn't find any theorems or anything to support this claim though. Possibly "For a real valued sequence {$s_n$}, $\lim_{n \rightarrow \infty}=s$ iff it's lim sup=lim inf =s (as n approaches infinity)" ??

 

[╔(σ_σ)╝]

The Attempt at a Solution

BUT, since {$a_n$} is bounded, it must be convergent?? i couldn't find any theorems or anything to support this claim though.

Well your claim is not quite correct; that is why you couldn't find a theorem to support it.

There are tons of bounded sequences that are not convergent.

Eg
(-1,1,-1,1...)

I have not tried the problem but I think I see what you mean by the sequence can be either decreasing or increasing.

I see 3 things
1) If the sequence is increasing then it is increasing at an increasing rate and is therefore not bounded.
2) If it is decreasing it is decreasing at a decreasing rate and could possible converge.
3) If it is constant then we know it converges.


I believe your best bet is to show that the sequence is cauchy. For one thing we know the sequence is not increasing. :)

 

[mynameisfunk]

is it possible to take $n$ to $-\infty$? We have never done that before..

 

[╔(σ_σ)╝]

is it possible to take $n$ to $-\infty$? We have never done that before..

Hey mate.
I made a mistake; b_n is actually increasing by definition. b_n is the distance between sucessive points and if you rearrange the first inequaility it is extremely easy to see that b_n-1 <= b_n which is what we want.

I actually arrived at that result in my original post but I was too stupid to notice that (1) was, in fact, the desired result.

Sometimes extremely obvious things are hard to see :-) :-).

 

[mynameisfunk]

Ya, I was able to arrive at that as well, but I am kind of confused by what I am supposed to do in the 2nd part. $a_n$ shouldn't converge since $a_n-a_{n-1} \leq a_{n+1}-a_n$, right?? But since $a_n$ is bounded, I suppose the $b_n$ must be bounded above?

 

[╔(σ_σ)╝]

The question ask you to either shown b_n is increasing or a_n is convergent. You have showed that b_n is increasing so you are done.

You do not have enough info to say that a_n converges. In fact I don't think it does.

 

[mynameisfunk]

OK, well what about this. Can i deduce from the fact that $a_n-a_{n-1} \leq a_{n+1}-a_n$ and that $a_n$ is bounded and is either monotonically increasing or decreasing and that every monotonically increasing/decreasing sequence converges iff it is bounded?

 

[╔(σ_σ)╝]

I don't think so but you should probably get a second opinion.

 

[l'Hôpital]

{a_n} can't converge. It's not Cauchy : ).

EDIT: Well, I guess if they were all negative terms, and b_n was increasing to zero, then maybe.

 

[mynameisfunk]

thanks, very helpful

 

[holomorphic]

{a_n} can't converge. It's not Cauchy : ).

EDIT: Well, I guess if they were all negative terms, and b_n was increasing to zero, then maybe.

b_n increases to zero is the key. That gives a_n convergent.