Is the Sequence {a_n} Convergent Given Its Recurrence Relation?

[MIT2014]

Homework Statement

For n$$\geq$$1 let 2a_n $$\leq$$ a_n-1 + a_n+1
Prove that a_n converges

Homework Equations

n/a

The Attempt at a Solution

2a_n+1 $$\leq$$ a_n + a_n+2
a_n+2 $$\geq$$ 2a_n+1 - a_n

How do I proceed? Ratio test?

 

[LCKurtz]

a_n = n would seem to be a counterexample.

 

[micromass]

Are you certain that this is correct?? Aren't there easy counterexamples?

 

[MIT2014]

I'm sorry. It should've been 2an instead of an. I've edited it above

 

[micromass]

What about the sequence

$$1,2,4,8,...,2^n,...$$

 

[LCKurtz]

I'm sorry. It should've been 2an instead of an. I've edited it above

Use the subscript button X₂ to do subscripts. Also the forum rules indicate you should not change a post that has already been replied to because it makes the discussion difficult to follow. Just correct it in the next post.

 

[CalTech>MIT]

First, Caltech>MIT lol.

Also, I believe you'd have to assume that the sequence a_n is bounded to solve this problem.

 

[MIT2014]

So how would you use boundedness to prove convergence?

PS. MIT>CalTech

 

[╔(σ_σ)╝]

From your sequence and with a little algebra it can be shown that
$$a_n - a_{n-1} \leq a_{n+1} - a_n$$.

Your sequence may converge if the different between sucessive terms approaches zero. If you don't impose some further restrictions I don't think you can show that this thing converges.

If a_n is bounded then things may work out.

 

[╔(σ_σ)╝]

I don't think you have enough information in this problem.

if $a_n - a_{n-1} \rightarrow 0$ then you have convergence.
But note that $a_n - a_{n-1}$ must be increasing to zero.

$$a_n$$ must be bounded but even more than that !

Edit
Something like ln(n) satisfies the condition of a_n - a_(n-1) increasing to zero but it does not converge since it is not bounded.

This question requires too many assumptions; something must be wrong with the question.

 

[CalTech>MIT]

I'm not very good at this stuff, so guys tell me if I am wrong somewhere, but here it goes:

2a_n $$\leq$$ a_n-1 + a_n+1
Through rearrangement: a_n - a_n+1 $$\leq$$ a_n-1 - a_n

This means that the difference between successive terms is decreasing. Since a_n is decreasing, the differences must decrease to 0 (this is where I'm concerned, can you make that assumption? otherwise how would you do it?). Thus, there exists N so for any n$$\geq$$N |a-a_n|<epsilon for any epsilon greater than 0. Thus, a_n must be a convergent sequence.

 

[micromass]

Hmm, I'm a bit skeptic. First of all, I see no reason to assume that it decreases to 0.
Secondly, take the sequence $$x_n=\sum_{k=1}^n{-\frac{1}{k}}$$. Then the difference between consecutive terms also decrease to 0. Still the sequence diverges...

 

[╔(σ_σ)╝]

I'm not very good at this stuff, so guys tell me if I am wrong somewhere, but here it goes:

2a_n $$\leq$$ a_n-1 + a_n+1
Through rearrangement: a_n - a_n+1 $$\leq$$ a_n-1 - a_n

This means that the difference between successive terms is decreasing. Since a_n is decreasing, the differences must decrease to 0 (this is where I'm concerned, can you make that assumption? otherwise how would you do it?). Thus, there exists N so for any n$$\geq$$N |a-a_n|<epsilon for any epsilon greater than 0. Thus, a_n must be a convergent sequence.

:-)
It doesn't have to be decreasing to zero. Infact, I has to be increasing to zero.

Just becase the difference approches zero doesn't guarantee convergence. In my above post I gave the examply of the sequence a_n=ln(n).