Prove that a convergent sequence is bounded

[s3a]

Homework Statement

The problem and solution are attached as TheProblemAndSolution.jpg.

Homework Equations

Definition of the limit of a sequence.

The Attempt at a Solution

I understand how P = ϵ + |A| can be seen as an upper bound that proves that the sequence is bounded, but for the last bit of the solution where the solution says “we choose P as the largest one of the numbers [. . .]”, why does it say a_N instead of a_n, shortly after? Is that a typo? Also, there isn't necessarily a value of a_n that is the largest (such that no value of a_n will ever reach ϵ + |A|), unless the author(s) meant to say that ϵ + |A| is one of the numbers to choose from such that ϵ + |A| is the largest number mentioned by the solution.

Could someone please clarify the last part of the solution for me?

I know this is a pedantic request, but I would greatly appreciate any input!

 

[STEMucator]

No that is not a typo.

$|a_n| < \epsilon + |A| = P$ for $n > N$.

By hypothesis you found an $N$ such that for any $n ≥ N, \space |a_n - A| < \epsilon$.

This means you can bound $|a_n|$ within $P$ for a sufficiently large $n$. This $P$ corresponds to the largest of the numbers $a_1, ..., a_N, (\epsilon + |A|)$ since we already found the $N$ such that $|a_n - A| < \epsilon$.

 

[s3a]

Oh, I think I get it now!

[1] I thought ϵ + |A| would always be the largest number, but if a_n = 1/n, for example, then a_1 would be the largest of the numbers at the end of the solution, right? Furthermore, it will always either be a_1 or ϵ + |A| that would be the largest number, and not a_2, . . ., a_(N–1), right?

[2] Though, since we're using n ≥ N (along with |a_n – A| < ϵ|) to define the limit (as can be seen by the fact that the numbers range from a_1 to a_N instead of a_1 to a_(N–1), at the end of the solution), that also means that |a_n| < ϵ + |A| for n ≥ N (instead of n > N), right?

I know this last minor detail makes no difference in the grand scheme of things, but I am a pedantic person. :P

 

[s3a]

Also, when the author(s) said to choose the largest one of the numbers for P, "largest", in this context, means with greatest magnitude, such that –1 is "larger" (in magnitude/absolute value) than +1/2, right?

 

[PeroK]

Also, when the author(s) said to choose the largest one of the numbers for P, "largest", in this context, means with greatest magnitude, such that –1 is "larger" (in magnitude/absolute value) than +1/2, right?

Hi s3a. That solution you posted is not very good. You've identified one problem. It really should be the largest of |a_1| ... |a_N|.

But, also, you're looking for P > |a_n|, so really it should be the largest of |a_1| ... |a_N| + 1.

Third, although it's sort of okay to leave ε as an arbitrary +ve number, it would seem better to me to take ε = 1, say, and find the value of N for which n > N => |a_n - A| < 1.

So, P = max{|a_1| ... |a_N|, |A|} + 1

 

[PeroK]

PS Don't forget to look for examples and not just rely on the analysis. E.g:

a_1 = 3, a_2 = 2, a_3 = 1, a_4 = 1/2, a_5 = 1/4 ...

In this case A = 0 and when ε = 1, we have N = 3:

n > 3 => |a_n - 0| < 1 => |a_n| < 1

P = max{|a_1|, |a_2|, |a_3|, |A|} + 1 = max{3, 2, 1, 0} + 1 = 4

 

[s3a]

Thanks for your answer, Perok.

Okay, so it seems that you're saying that P = max{|a_1|, |a_2|, . . ., |a_N|, |A|} + ϵ.

However, it seems to me that |a_1|, |a_2|, . . ., |a_N| each already have an error margin that is greater than or equal to ϵ, with respect to |A| (instead of strictly less than ϵ, because, for the error margin to be strictly less than ϵ, n must be greater than N), so, from that, it seems to me like it should be P = max{|a_1|, |a_2|, . . ., |a_N|}, since it seems to me that |a_i| ≥ |A| + ϵ, where i is any number from [1, N] (or [0, N], if a_0 is defined).

If I'm wrong, could you please give me an example where |a_i| < |A| + ϵ?

 

[PeroK]

Thanks for your answer, Perok.

Okay, so it seems that you're saying that P = max{|a_1|, |a_2|, . . ., |a_N|, |A|} + ϵ.

However, it seems to me that |a_1|, |a_2|, . . ., |a_N| each already have an error margin that is greater than or equal to ϵ, with respect to |A| (instead of strictly less than ϵ, because, for the error margin to be strictly less than ϵ, n must be greater than N), so, from that, it seems to me like it should be P = max{|a_1|, |a_2|, . . ., |a_N|}, since it seems to me that |a_i| ≥ |A| + ϵ, where i is any number from [1, N] (or [0, N], if a_0 is defined).

If I'm wrong, could you please give me an example where |a_i| < |A| + ϵ?

a_1 = 0

 

[s3a]

a_1 = 0

Thanks for that.

What about a sequence formula, where there is at least one nonzero a_i such that |a_i| < |A| + ϵ?

Would you be able to give me such an example, by any chance? :)

 

[PeroK]

Thanks for that.

What about a sequence formula, where there is at least one nonzero a_i such that |a_i| < |A| + ϵ?

Would you be able to give me such an example, by any chance? :)

$a_n = 1 - (\frac{1}{2})^n$

Don't get hung up on formulas. The mathematical properties of a thing do not depend on expressing that thing as a formula.

 

[LCKurtz]

Thanks for that.

What about a sequence formula, where there is at least one nonzero a_i such that |a_i| < |A| + ϵ?

Would you be able to give me such an example, by any chance? :)

Would you be able to come up with an example yourself? Have you tried?

 

[s3a]

Thanks again, Perok.

Okay, so for n ≤ N, a_n = 1 – (1/2)^n seems to be a formula for which |a_i| < |A| + ϵ.

It also seems that, for n ≤ N, a_n = (1/2)^n is a formula for which |a_i| ≥ |A| + ϵ.

About the formulas, I just needed the formulas to build my intuition.

And, about finding one myself, I think I was a little too confused with everything at the time.

I see now that the a_1, a_2, . . ., a_N shouldn't be a_1, a_2, . . ., a_n, because for n > N, it would always be the case that |a_n| < |A| + ϵ and |A| + ϵ is one of the candidates for the value of P, and it's pointless to add candidates for P that are guaranteed to not be the largest.

I have one last confirmatory question, shown below.:
Since there are two definitions for the set of natural numbers (one including 0 and another excluding 0), if we used the definition of natural numbers including 0, we would have a_0, a_1, a_2, . . ., a_N instead of a_1, a_2, . . ., a_N as the candidates for P (in addition to |A| + ϵ), right?

 

[Fredrik]

Since a complete solution has already been given, I will offer my version:

The key to this problem is the realization that "almost all" terms are close to A, which is equal to |A| or -|A|, depending on whether A is negative or non-negative.

Suppose that $\lim_n a_n =A$. Then every open interval with A at the center contains all but a finite number of the $a_n$. In particular, the open interval $(A-1,A+1)$ contains all but a finite number of the $a_n$. We don't know if A is negative or non-negative, but in both cases, the open interval $(A-1,A+1)$ is a subset of the closed interval $[-|A|-1,|A|+1]$. Let N be a positive integer such that $a_n\in [-|A|-1,|A|+1]$ for all $n\geq N$. Some of the first N-1 terms may be outside of the interval [-|A|-1,|A|+1], so we can't just set P equal to |A|+1. But if we set $P=\max\{|a_1|,\dots,|a_{N-1}|,|A|+1\}$, the interval [-P,P] will contain all terms.