Prove that a convergent sequence is bounded

1. Aug 20, 2014

s3a

1. The problem statement, all variables and given/known data
The problem and solution are attached as TheProblemAndSolution.jpg.

2. Relevant equations
Definition of the limit of a sequence.

3. 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!

Attached Files:

• TheProblemAndSolution.jpg
File size:
21.4 KB
Views:
134
2. Aug 20, 2014

Zondrina

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 $|a_n - A| < \epsilon$.

3. Aug 20, 2014

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

4. Aug 20, 2014

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?

5. Aug 20, 2014

PeroK

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

Last edited: Aug 20, 2014
6. Aug 20, 2014

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

7. Aug 21, 2014

s3a

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?

8. Aug 21, 2014

PeroK

a_1 = 0

9. Aug 21, 2014

s3a

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? :)

10. Aug 21, 2014

PeroK

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

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

11. Aug 21, 2014

LCKurtz

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

12. Aug 23, 2014

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?

13. Aug 23, 2014

Fredrik

Staff Emeritus
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.

Last edited: Aug 23, 2014