Need help understanding something in my proof

Samuelb88

1. The problem statement, all variables and given/known data
Suppose [tex]S_n > 0[/tex] and [tex]r S_n > S_{n+1}[/tex], where r is a constant such that 0<r<1. Show that [tex] \lim_{n\rightarrow +\infty} {S_n} = 0[/tex].

3. The attempt at a solution

First, I'd like to know if my proof is correct.

Proof: Suppose [tex]S_n > 0[/tex] and [tex]r S_n > S_{n+1}[/tex], where r is a constant such that 0<r<1. First note [tex]S_n[/tex] is decreasing since [tex]S_n > S_{n+1}[/tex] if [tex]r S_n > S_{n+1}[/tex]. Moreover, for every n [tex]S_n > 0[/tex], so [tex]r S_n > S_{n+1} > 0[/tex]. Therefore the [tex] \lim_{n\rightarrow +\infty} {S_n} = 0[/tex].

Suppose my proof is correct. Then my question is what role does r play in this problem? It seems to me, so long as [tex]S_n[/tex] is decreasing, and [tex]S_n > S_{n+1}[/tex], and for every n [tex]S_n > 0[/tex], then the [tex] \lim_{n\rightarrow +\infty} {S_n} = 0[/tex]. Right?

╔(σ_σ)╝

Your proof is not correct, S_n could be decreasing but the limit could non-zero. The fact that it is bounded below by zero doesn't imply it approaches zero. Example: consider the sequence 1/2 + 1/n. Cleary, it is decreasing and bounded below by zero but the limit is 1/2.

I think you have to show that r^n goes to zero and since S_n is bounded then the sequence S_n goes to zero since we can make S_n as small as we like.

Fisicks

The problem says for all n rSn>Sn+1. This isnt a proof but it leads to it. Say r is .5 for example. Then .5 > Sn+1/Sn. It seems easy to derive a contradiction when lim of Sn is non zero. If you keep taking halves of Sn, you will go lower than your non zero limit. But if Sn is approaching zero, then you can easily keep taking halves and never even touch zero. Now just generalize.

Samuelb88

I am still a little confused about how to conclude that the [tex] \lim_{n\rightarrow +\infty} {S_n} = 0[/tex]. Here is what I've come up with:

Suppose for every n, [tex]S_n > 0 [/tex] and for some fixed r, 0<r<1, [tex]r S_n < S_{n+1}[/tex]. Let r=1/(1+a), a > 0. Then [tex]S_n[/tex] is a decreasing sequence, and is bounded below. Then for every n

[tex] 0 < \frac{S_{n+1}}{S_n} < \frac{1}{1+a}[/tex].

We can arbitrarily make 1/(1+a) as close to zero as we want by choosing an extremely large a, and the inequality will still hold. In other words, I can squeeze [tex]S_{n+1} / S_n[/tex] between 0 and 1/(1+a) << 1.

At this point, may I conclude [tex] \lim_{n\rightarrow +\infty} {S_n} = 0 [/tex] since the ratio of [tex]S_{n+1}[/tex] and [tex]S_n[/tex] can be squeezed between 0 and a very small number, the sequence [tex]S_n[/tex] is decreasing and is bounded below by zero?

Fisicks

The point is, Sn will go to zero if r is .99999999 or .00000001 and you cannot choose r, it is some given value between 0 and 1. Think about what (r=.5) .5>Sn+1/Sn means. It means at the least, Sn+1 is 2 times smaller than Sn+1 FOR ALL n.

l'Hôpital

Try finding a bound for S_n in terms of n .(Hint: Try induction.)

╔(σ_σ)╝

The way I envisioned the proof was like this

[tex] rS_0 > S_1[/tex]
[tex] r^{2}S_0 > S_2[/tex]
.
.
.
[tex] r^{n}S_{0} > S_{n}[/tex]

Using the definition of a limit we want to show that [itex] S_n \rightarrow 0[/itex]

[tex] 0< | S_{n} -0 | < |S_{0}|r^{n} [/tex] . We know [itex] S_0[/itex] is bounded since the original sequence is convergent.

If you have already proving that [itex] \lim_ {n \to \infty } r^{n} = 0 \quad 0<r<1[/itex] then the proof is complete.

Samuelb88

Okay, for the most part, I understand what you are saying here. (Didn't understand how you derived the relation between [tex]r^n S_1[/tex] and [tex] S_n[/tex] originally... :)

How about this:

We want to show using the definition of a limit that [tex] S_n \rightarrow 0 [/tex] if for ever E >0, there is an N s.t. whenever n>N, [tex] |S_n| < E[/tex]. Let E >0. Suppose [tex]S_n > 0 [/tex] (bounded) and [tex] r S_n > S_{n+1} [/tex] (decreasing). So

[tex] rS_1 > S_2[/tex]
[tex] r^{2}S_1 > S_3[/tex]
.
.
.
[tex] r^{n}S_{1} > r^{n-1} S_{1} > S_{n}[/tex] which is true for all n.

Let [tex] E= r^{n} S_1[/tex]. As required. Since [tex]S_n < r^n S_1[/tex] for all n, and [tex]r^n \rightarrow 0[/tex] as [tex] n \rightarrow +\infty[/tex], then the [tex] \lim_{n\rightarrow _\infty} {S_n} = 0[/tex]. ?

╔(σ_σ)╝

No, you can't just set [tex] \epsilon = r^{n} S_0 [/tex]

The definition of a limit requires that for [tex] n > n_{0} [/tex] the limit holds and [tex] n_0[/tex] should depend of epsilon.

Also [tex] 0 < S_n < r^n S_1[/tex] the 0 < part is important since you want [tex] S_n[/tex] to be sandwiched by zero from the left and right.
You want to pick an [tex] n_0[/tex] such that for [tex] n > n_{0} [/tex] you can make tex] r^n S_1[/tex] as small as possible.

I am not sure if you have to prove this fact or you can just use it .

If you are simply allowed to use it then you can just state that since r^n goes to zero then epsilon and n_0 exist. Or you could jsimply use the squeeze theorem/ sandwich lemma.

Samuelb88

Okay, so we agree that for every n, [tex] 0 < S_n < r^{n} S_{1} [/tex]. And from my book, there is an example which proves that for every a such that 0<a<1, the [tex] \lim_{n\rightarrow +\infty} {a^n} = 0[/tex]. So I may use that fact in my proof. I was thinking that since we know when 0<r<1 that [tex]r^n \rightarrow 0[/tex] as [tex] n \rightarrow +\infty[/tex], then we know for every ε > 0, there must exist an N such that whenever n > N, [tex] | r^n | < \epsilon[/tex]. So couldn't we just arbitrarily choose an n > N (since we know such N exists) such that [tex] S_1 |r^n| < \epsilon[/tex]. Since 0<r<1, and for every n, [tex] S_1 > 0[/tex], then [tex] S_1 |r^n| = S_1 r^n[/tex], and therefore we have found such n such that [tex] 0 < S_n < r^n S_1 < \epsilon [/tex].



I am pretty sure I can not use the fact that [tex] \lim_{n\rightarrow +\infty} {S_n} = 0[/tex], that is, that's what we're trying to prove. At any rate, I was thinking about using the squeeze theorem, but I thought the squeeze theorem required less than or equal to signs, and we only have less than signs in our inequality. Other wise I would be squeezing the limit of [tex] S_n [/tex] between some greater than zero, and less than zero, which to me seems invalid. Can you clarify this for me, please?

HallsofIvy

You can use the fact that [itex]0< S_n< r^nS_1[/itex], to argue that [itex]|S_n- 0|= |S_n|< |S_1|r^n< \epsilon[/itex] implies that [itex]r^n< \epsilon/|S_1|[/itex]. How large does n have to be in order that that be true?

Samuelb88

Great. I would have to choose n to be an integer greater than [tex] ln ( \frac{( \epsilon }{|S_1| }) / ln(r)[/tex]. This choice for n makes sense because 0<r<1, ε > 0, and [tex] |S_1| > 0 [/tex].

Edit: Oops, forgot to take ln on other side. :D

╔(σ_σ)╝

Good. You are basically done. Just everything together.

╔(σ_σ)╝

Btw your inequality is wrong.
[tex] S_n < r^{n}S_0[/tex] is correct if you start from n=0.

But since you are starting for n=1 your inequality should be...
[tex] S_n < r^{n-1}S_1 [/tex].

You made a mistake when you said.... [tex] r^{n-1}S_1 < r^{n}S_1[/tex]. It is not true since 0< r<1.

This is just a minor correction since [tex] r^{n-1} \rightarrow 0[/tex].

Samuelb88

Suppose [tex] 0 < S_n[/tex] and [tex] S_{n+1} < r S_n [/tex] where r is a constant such taht 0<r<1. Then [tex] S_n < r^{n} S_0[/tex].

Lemma: From an example in my book, we know when 0<a<1 that the [tex]lim_{n\rightarrow +\infty} {a^n} = 0[/tex]. That is, for every ε > 0, there is an N such that whenever n>N, [tex] |a^n| < \epsilon[/tex].

Note that the sequence is bounded below, since for every n, [tex] 0 < S_n[/tex]. Then for every n, [tex] 0 < S_n < S_0 r^n [/tex] which implies for some n>N, [tex] 0 < |S_n| < |S_0| r^n < \epsilon [/tex], which implies [tex] r^n < \epsilon/|S_0|[/tex]. Observe that [tex]0 < r, \epsilon, |S_0|[/tex], so take [tex] N = ln( \epsilon/|S_0|) / lnr [/tex], and choose n>N. Then [tex]|S_n - 0| < \epsilon[/tex] and therefore [tex] lim_{n\rightarrow +\infty} {S_n} = 0[/tex]. Done!

How does this look?

╔(σ_σ)╝

This part is not too good.

This part is not true.

What is true is that ....
[tex] \lim_{n \to \infty} r^{n} = 0[/tex] implies that there exist [tex]N_0[/tex] such that for [tex] n > N_{0}[/tex],[itex] r^{n} < \frac{\epsilon}{|S_0|}[/itex].


The rest of the proof is okay.

Samuelb88

Okay, now I see how I am suppose to prove the limit is zero. Thanks much, all of you! :)