Proof: Convergent Sequence is Bounded | Real Analysis

[@Nate3145-zt8rh]

I dont care how many times you have done this proof, if you have even a slight difficulty recalling it, do it again. It is so fundamental to analysis.

[@Nate3145-zt8rh]

what i mean is that this proof should be intuitive and natural

[@kunalkumar-ob4lg]

The explanation cannot be more simple and easy to understand. Thank you.

[@WrathofMath]

So glad it helped, thanks for watching and check out my real analysis playlist if you're looking for more! kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli

[@kunalkumar-ob4lg]

@@WrathofMath The playlist is fire. It summarizes my whole course. I'm definitely gonna watch that.

[@danfirth3017]

these videos have carried me through this first term of analysis

[@agrani7650]

let's just say that you are my professor for the first semester ... LOVE YOU and YOUR HARDWORK ! Thank you

[@WrathofMath]

Glad to help! Thanks for watching! If you're looking for more analysis, check out my playlist! kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli

[@abdellahomaradolf3566]

Your carefully thought through explanation is superb. Whenever I learn a concept from your explanations, no matter how abstract it is, I concretely understand the concept.

[@WrathofMath]

That's awesome to hear, thank you for watching!

[@denglannieee6702]

OMG, this way is so much simpler than my lecture. U r sooo smart! Thank you so much.

[@WrathofMath]

I'm so glad to help, thanks for watching!

[@joshbolton2782]

Dude you are so good at explaining the proof so it makes sense. Thank you man.

[@Bedoroski]

Thanks for so much hard work into explaining it. The textbook by Jim Cummings somehow made it a bit confusing with the inequalities

[@WrathofMath]

Glad to help, thanks for watching!

[@siriuss_]

Thanks for the video! Also, I'm trying to show that a cauchy sequence then it is bounded (similar to the video) we have, if epsilon= 1, then |a(n) - a(k) < 1. This is similar to -1 < a(n) - a(k) < 1 => a(k) -1 < a(n) + 1. Also, -(a(k) + 1) < a(k) - 1. So we have, -(a(k) + 1) < an < a(k) + 1.If, n = {1,..,N} then a(k) ≥ a(N), so we have (Taking, a(k) + 1 = K), -K ≤ a(1),..a(N) ≤ K. Now, taking M = K + 1, we have -M < a(1),..a(N) < M. Then my question is that i know that K = max{a(1),...,a(N), 1 + a(k)). So what is M then? I mean in max{} or something like that?

[@siriuss_]

This was a proof in Charles Pugh' Real analysis. I tried to fill in my details myself but now I'm confused.

[@siriuss_]

Another proof that I saw in abbot's Analysis. We know that, |a(n) - a(k)| < 1, but also, a(n) ≥ a(N), so if we take k = {1,..,N}. So, |a(n)| ≥ |a(N)| then, |a(n)| > |a(N)| + 1. So, If |a(n)| > M , then we take, M = max{1,...,|a(N - 1)|, |a(N)| + 1}. Not |a(N)| since, |a(n)| ≥ |a(N)|. Thus we have shown that the sequence is bounded. Can you improve on my reasoning here?

[@cloudhuang6691]

Sir, your lesson is very comprehensive, thank you so much.

[@WrathofMath]

Glad to help - thanks for watching!

[@cloudhuang6691]

@@WrathofMath Excuse sir, I've reconsidered this basic question again recently, and something occured to me: Is'nt there any convergent sequence blows up divergently in some of the first few "N" terms so that makes the sequence unbounded?

[@turokg1578]

this channel is a lifesaver...

[@WrathofMath]

So glad to help! I'm continuing to work on the analysis playlist, let me know if you have any questions!

[@turokg1578]

@@WrathofMath is it an acceptable proof if i take M = max {|a_1|, |a_2|, ... |a_N|, |1+a|, |a-1|} and say that |a_n|

[@jonhroldandl.buenaventura7596]

Can I use ε itself instead of assigning specific value? since it is defined to be greater than 0. Thank you!

[@SoumaHsu]

Very clear explanation! Helps me a lot

[@maxpercer7119]

For those using the alternate definition of bounded sequence: "a sequence (a_n) is bounded if |a_n |

[@federjcoo]

doesnt this work only for epsilon=1? like what if i choose epsilon=2?

[@chk374]

Good video, thank you. Question: When arriving at the end result a_k

[@WrathofMath]

Thanks for watching and great question! We know a_1,a_2,...,a_N are bounded because N is a fixed natural number. In other words, the list is finite. Remember that N comes from a_n being convergent, it is the N such that all terms of the sequence after the Nth term are within 1 of epsilon, guaranteed to exist by convergence. Does that help?

[@chk374]

@@WrathofMath The question is more aimed at showing that a_1,a_2,…,a_N are bounded (i understand a_1,a_2,…,a_N is a set with a finite number of elements but that doesn’t guarantee said elements are bounded), ie how do you know that a_j

[@WrathofMath]

I'm not sure I understand. The set being finite does guarantee it's bounded, since a finite set has a max and a min, which are bounds. We are talking about sequences of real numbers, so certainly every a_j is less than infinity not only for j = 1,2,...N, but for all j. Regarding 1/|k-2| I'm also not sure what you're pointing out. When you say not all a_k are bounded do you mean each a_k is not necessarily bounded? That wouldn't make sense since each a_k is just a number. So do you mean the set of all a_k for positive integers k not equal to 2? Maybe you're saying "How do we know none of our a_n are something like 1/0?" The answer would be that a_n was taken to be a sequence of real numbers.

[@chk374]

@@WrathofMath What if my sequence is 1/(k-2)? a_2 is unbounded. Doesn’t boundedness mean a_k < oo for all k? What am i missing here

[@chk374]

is the answer here that k in Narural# implies that a_2 is not defined (and not oo) and therefore a_k is really only defined for k>2?

[@gmingwll7985]

Thanks for this deep concept sir.

[@BubimedesOfficial]

Do you have to select an epsilon? Because the way I understand it, it wouldn't make a difference (since you're keeping everything else abstract)

[@anveshshrivastava2115]

Literally, cant get a better expalination

[@xssavage7104]

I'm only finding out about this channel 2 hours before an analysis exam... it's already late

[@Kira-cw1yp]

That was such an amazing explanation.... Thanks buddy

[@WrathofMath]

Glad to help, thanks for watching!

[@Kira-cw1yp]

@@WrathofMathI am first year student at IIT for undergrad in computer science

[@WrathofMath]

That's awesome - good luck, computer science is a fascinating pursuit!

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Thank you for the great work , i will be pleased to hear the optimal way to achieve your level in maths .

[@scp_at_iitb]

Very nicely explained 👍

[@WrathofMath]

Thank you!

[@이주홍-d5u]

good and clear proof!

[@WrathofMath]

Thank you!

[@ezenaboifeyinwa]

When u were explaining for n = 1,2,3,...,n u said that it's possible for those n to be bigger than 1+a please I don't understand how 🙏

[@TheThBe]

Doesn't it suffice to prove that |An| =< U, since it will also be bounded by -U? Excellent video nonetheless, as always!

[@avtaras]

Might be a dumb question, but just because one can find a value ε = 1 such that ∀n>N, |xₙ − a| < ε, it doesn’t mean that the sequence converges? If one took epsilon to be arbitrary then it would make sense. So how can one just say epsilon is equal to 1 and proceed with the proof.

[@WrathofMath]

Thanks for watching and the question! It's because the sequence's convergence is not in question. We are assuming the sequence is convergent, and thus the for every n>N stuff must be true for ANY epsilon, so we are certainly at liberty to take a specific epsilon which is convenient for our proof, that the sequence is bounded.

[@blackhole1222]

thanks! @@WrathofMath

[@Dupamine]

I have the same question as @ezanabo < When u were explaining for n = 1,2,3,...,n u said that it's possible for those n to be bigger than 1+a please I don't understand how 🙏

[@WrathofMath]

Can you give me a timestamp?

[@Dupamine]

@@WrathofMath It is at 1:30. a-1

[@WrathofMath]

{1,2,3,4,5} is not a sequence because a sequence is a function from the naturals to the reals, which is to say it has infinitely many terms. If a sequence converges, as we've assumed ours converges to a, then for any epsilon>0, there is some point in the sequence a_N, after which all terms of the sequences are within epsilon of the limit a. So if epsilon = 1 for example, there is some a_N after which |a_n - a| < 1 for all n>N, which means -1 < a_n - a < 1 for all n>N. Since the set you mention is not a sequence, and so it certainly doesn't converge, this line of reasoning doesn't apply to it. Does that help?

[@WrathofMath]

On the other hand, if we consider the sequence 1, 2, 3, 4, 5, 5, 5, 5, ... Then it is true that 5 - 1 < a_n < 1 + 5 for all n>4. After this point, the sequence is within 1 of its limit, and in particular for this sequence it happens to equal its limit after this point. Remember the definition of convergence guarantees this after some point in the sequence, it certainly need not be true for every term of the sequence.

[@sanasunu7]

Tnx for ur classs.. its helpful for ma xam

[@WrathofMath]

So glad to help! Thanks for watching and check out my analysis playlist if you're looking for more: kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli

[@sanasunu7]

@@WrathofMath yesterday i complete my real analysis xam..its too easy.. bcz i watch ur classes

[@jingyiwang5113]

Your explanation is really clear. It is so helpful for me~Thank you so much!

[@WrathofMath]

So glad to help, thanks for watching! Let me know if you have any questions!

[@Puzzlers100]

Can a sequence be undefined at a point but still converge? 1/(n-1) would make a sequence where the first term is undefined (divison by zero) and the remaining terms would eventually converge to 0. Is this not a valid sequence?

[@WrathofMath]

Good question, I would just consider this sloppily written mathematics. A sequence is a function from the naturals to the reals, meaning there is a term assigned to position 1, 2, 3, etc. In the case you describe, the "undefined term" simply isn't a term in the sequence, and so really the sequence should be 1/n, since this describes the intended sequence without the weird ambiguity of a first "undefined" term. A sequence is a function, so it can't have any undefined values, since a function has one defined output for each input.

[@hussainfawzer]

What’s the software you’re using here ?

[@WrathofMath]

Notability for iPad! It's a wonderful app.

[@mathandsciencetips]

Good explaination

[@WrathofMath]

Thanks, glad it was clear! Let me know if you ever have any video requests, and if you're looking for more analysis - check out my playlist: kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli

[@munugotianusha5305]

Nice teaching 👌👌

[@WrathofMath]

Thank you! Let me know if you have any questions, and check out my analysis playlist for more! kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli

[@tiyabanda7555]

Clear explanation

[@WrathofMath]

Thank you!

[@mahmoudalbahar1641]

Thanks.

[@WrathofMath]

Thank you for watching!

[@daudierastomlangi2218]

Thanks

[@chin6796]

Thanks.

[@PAWAN-SINGH-IIT-DELHI]

Awesome

[@WrathofMath]

Thank you!