# Show Convergence of sequence (1 + c)(1 + c^2) (1 + c^n)

1. Jan 24, 2012

### looserlama

1. The problem statement, all variables and given/known data

Hey, so here is the problem:

Suppose 0≤c<1 and let an = (1 + c)(1 + c2)...(1 + cn) for integer n≥1. Show that this sequence is convergent.

Well I understand the basic concepts of proving convergence of sequences, but in class we've only ever done it with sequences where the terms are summed not multiplied like this one is. I guess I'm just having trouble figuring out where to go really...

2. Relevant equations

I guessed we either need to use the theorem that says any bounded monotone sequence is convergent.

Or by simply finding the limit L and proving that the sequence converges to L.

3. The attempt at a solution

There's a suggestion that says " Show that 1 + c ≤ ec ".

I did that by showing that if c = 0, then 1 + 0 = e0 = 1

And then for any c>0 d/dc(1 + c) < d/dc(ec).

Therefore for any c≥0, 1 + c ≤ ec.

Now I don't really know how to use this to solve my problem.

I tried to show that the sequence is monotone:

As an+1 = an(1+cn+1)

Therefore an+1 ≥ an

Therefore an is non-decreasing, therefore monotone.

But now I have no idea what to do. I really need help cause my prof just says that it should be obvious.

Last edited: Jan 24, 2012
2. Jan 24, 2012

### tiny-tim

welcome to pf!

hi looserlama! welcome to pf!
hint: 1 + c2 ≤ … ?

3. Jan 24, 2012

### looserlama

thanks tiny-tim.

Would it not be 1 + c2 ≤ ec2 ?

Does that just mean (1 + c)(1 + c2)...(1 + cn) ≤ ecec2...ecn = ec + c2 + ... + cn = e(c - cn+1)/(1 - c)

That's probably taking it too far, I just don't really understand where to go with that?

4. Jan 24, 2012

### Dick

That's not taking it too far. Take it even farther. e^((c - c^(n+1))/(1 - c)))<=e^(c/(1-c)). So your products are bounded from above. Now what?

5. Jan 24, 2012

### looserlama

Oh yea, so if an is smaller than a finite number, then it's bounded from above and given an ≥ 0, it's bounded. Therefore it's convergent.

But I need to prove that 1 + cn ≤ ecn for any integer n ≥ 1 .

I don't know how to do that?

The only way I can think of is with induction, so this is what I have:

Basic Step: For n = 0

1 + c ≤ ec

Inductive Step:

If the statement holds for integer k ≥ 1, then

1 + ck ≤ eck

so 1 + ck + 1 ≤ ceck

but how do I show that ceck ≤ eck + 1 ?

Last edited: Jan 24, 2012
6. Jan 24, 2012

### Dick

Now you are deviating from the correct path. Sure, an<=e^(nc). But that's no use. As n->infinity e^(nc) goes to infinity. Bad. And proving 1+c^n<e^(c^n) is no real trick if you've already proved 1+c<e^c. Just substitute c^n for c!

7. Jan 24, 2012

### looserlama

Yea I saw that what I wrote before was wrong, so I edited my previous post.

I just am having trouble rigorously proving that 1 + c^2 <= e^(c^n). See my edited post for what I did.

8. Jan 24, 2012

### Dick

If 1+c<=e^(c) then 1+c^2<=e^(c^2). Just substitute c^2 for c in the first equation. I told you that. You don't need a separate proof if you already proved the first one.

9. Jan 25, 2012

### looserlama

Oh I see. U was thinking about that the wrong way.

Thanks Dick