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

[looserlama]

Homework Statement

Hey, so here is the problem:

Suppose 0≤c<1 and let a_n = (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...

Homework 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.

The Attempt at a Solution

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

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(e^c).

Therefore for any c≥0, 1 + c ≤ e^c.

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

I tried to show that the sequence is monotone:

As a_n+1 = a_n(1+cⁿ⁺¹)

Therefore a_n+1 ≥ a_n

Therefore a_n 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.

 

[tiny-tim]

welcome to pf!

hi looserlama! welcome to pf!

There's a suggestion that says " Show that 1 + c ≤ e^c ".
…
Now I don't really know how to use this to solve my problem.

hint: 1 + c² ≤ … ?

 

[looserlama]

thanks tiny-tim.

Would it not be 1 + c² ≤ e^c2 ?

Does that just mean (1 + c)(1 + c²)...(1 + cⁿ) ≤ e^ce^c2...e^cn = e^{c + 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?

 

[Dick]

thanks tiny-tim.

Would it not be 1 + c² ≤ e^c2 ?

Does that just mean (1 + c)(1 + c²)...(1 + cⁿ) ≤ e^ce^c2...e^cn = e^{c + 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?

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?

 

[looserlama]

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

But I need to prove that 1 + cⁿ ≤ e^cn 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 ≤ e^c

Inductive Step:

If the statement holds for integer k ≥ 1, then

1 + c^k ≤ e^ck

so 1 + c^{k + 1} ≤ ce^ck

but how do I show that ce^ck ≤ e^{ck + 1} ?

 

[Dick]

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

But I need to prove that 1 + cⁿ ≤ e^cn.

I think I found a way to show that a + cⁿ ≤ e^c for any integer n ≥ 1, this is what I did (not sure it's right though):

0 ≤ c < 1 , therefore c = 1/b for b > 1.

So we want to show that 1 + 1/bⁿ ≤ e^1/b

-We already know this statement is true if n = 1 (cause I already showed that)

-For any n > 1 :

d/dn(1 + 1/bⁿ) = -n/b^n-1

d/dn(e^1/b) = 0

So -n/b^n-1 < 0

Therefore 1 + 1/bⁿ ≤ e^1/b

So 1 + cⁿ ≤ e^c for 0 ≤ c < 1


Is that a rigorously correct proof?
(I guess I could just use induction instead, given it's just for integers n)

So changing the solution using this, we get:

a_n = (1 + c)(1 + c²)...(1 + cⁿ) ≤ e^ce^c...e^c = e^{c + c +...+ c} = e^nc which is a finite number.

Therefore a_n is bounded, so a_n is convergent.

That makes sense right?

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!

 

[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.

 

[Dick]

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.

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.

 

[looserlama]

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

Thanks Dick