Real Analysis 3 | Bounded Sequences and Unique Limits

Please consider to support me if this video was helpful such that I can continue to produce them :)

🙏 Thanks to all supporters! They are mentioned in the credits of the video :)

This is my video series about Real Analysis. We talk about sequences, series, continuous functions, differentiable functions, and integral. I hope that it will help everyone who wants to learn about it.

This is part 3 of 64 videos. Let's discuss some bounded sequences.

00:00 Intro
00:15 An example for showing that a sequence is divergent
03:41 Definition of a bounded sequence
04:36 A convergent sequence is also bounded
06:09 The limit of a convergent sequence is uniquely given
08:59 Outro

#RealAnalysis
#Mathematics
#Calculus
#LearnMath
#Integrals
#Derivatives

I hope that this helps students, pupils and others. Have fun!

(This explanation fits to lectures for students in their first and second year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[chilledvibes]

i like how you describe everything so simply. my uni lecturers usually just rush through the small details so u never really grasp the full thing

[punditgi]

Another excellent video. This series is truly a gem! 😃

[suryanarayanachebolu]

Thanks for the Valuable Lecture. Entire Math from Degree to P.G has a lot to do with "Real Analysis" .

[ChrisOffner]

I don't understand why you choose *epsilon = 1* here, and why then *|a_N - a| < epsilon* and *|a_(N+1) - a| < epsilon* are supposed to hold. If *a = 0, * then neither of those hold, because the distance between *a_N* and *a = 0, * as well as the distance between *a_(N+1)* and *a = 0, * are both equal to *epsilon = 1, * not less than *epsilon = 1.* What am I missing here?
My confusion starts basically at ~1:42 when you say "We have to choose *epsilon* so small that in the *epsilon* neighbourhood around *a, * we don't have *-1* and *1* at the same time. This means we need a number smaller or equal than *1.* "
Why a number *equal* to *1* ? Doesn't this only work with numbers *smaller* than *1* (see case *a = 0* )?

[covariance]

Hey! Just wanted to say thanks so much for these excellently made videos! I'm, uh, kind of a screw up in life (did a bachelor's and master's in psych; dropped out of a phd psych program after many years... returning to school as a mature student) and am planning on taking Real Analysis I in January 2024 (*knocks on wood that nothing unexpected comes up to stop me from doing that*) and I've heard lots of stories about how difficult this course/topic tends to be.

Thus, I am very grateful to be able to see your videos! I don't have a lot of money at the moment, but if things go swimmingly, I will definitely say thanks with the "Tip" button on youtube. Or actually, I see the Paypal link in the youtube description; perhaps I'll use that instead! :)

Anyway, I hope to continue watching this playlist and making notes. In the meantime, I just wanted to express my gratitude in case I may forget later.

On topic: I definitely would not have thought to use the triangle inequality to prove the diveregence of (-1)^n; I never even thought to formally prove it, before! I've only had a little experience with epsilon-delta proofs in Calc II (but obviously not enough). Will be taking Calc III and a discrete math (they call the course mathematical structures at my school) in the fall prior to real analysis in the winter, so fingers crossed I'll be a little less inept by then :)

[atlanta]

Thank you for the videos:) These help me understand topics so much better than my uni lectures

[priyamhazarika]

Very helpful . Understanding everything vry clearly . Thank you

[luissalasar]

Congrats on the excellent lecture! Could u tell me what software you use to write on the screen?

[triton]

Do we have to use the TI in the end there or is it enough to say mod(a-ā)<2ε because it is implied by the definitions of convergence

[hyperduality]

Convergence (syntropy, homology) is dual to divergence (entropy, co-homology) -- the 4th law of thermodynamics!
Bounded is dual to unbounded, closed is dual to open.
Elliptic curves are dual to modular forms.
"Always two there are" -- Yoda.

[IITIAN_ROHIT]

well understood and I'm in IIT DELHI the professor here is not understanding me . anything if it is sufficient for engineering maths then I can follow your lecture what is proper
please tell me SIR
🙏🙏🙏

[maxpercer]

For the proof of ' (a_n) convergent → (a_n) is bounded ' at 6 minutes,
I would just add, you have to pick a specific epsilon. A convenient one is, e = 1.
Or if you prefer to leave it as epsilon, without specifying its length, call it epsilon*, or epsilon_0.

[jaydensidabutar]

hello, thank you for putting out these fantastic mathematics series! these help me understand topics like analysis and calculus so much better,
and just a quick question though which sort of went over my head, why is it that the part in 8:40 can't happen?

[tatasharada]

Thanks for this excellent course material. Many many thanks for all the efforts. I have a question. If I have a sequence an = 1/ (n-1); clearly this sequence is convergent because as n goes to infinity the sequence goes to zero. But will it be considered bounded? Because at n = 1, it starts off at 1/0 ~ infinity? How would that be different from a sequence say an = n^2 which is divergent. So in that vein is the sentence all convergent sequences are bounded valid?

[ThreeMiningHD]

Hi, du machst echt coole Videos! Ich wollte Mal fragen, ob du auch über Übungszettel zu deinen verschiedenen Videoreihen verfügst? LG

[jamesyeung]

what's the difference between ε-N proofs and infinitesimal proofs? isn't ε also getting "infinitely close to zero"?

[PunmasterSTP]

Bounded sequences? More like “Boredom is left in pieces!” Thanks for another stimulating lecture.

[angusclark]

Hello! I am slightly confused about convergence implies bounded. In your example, a1 and a2 are both outside the epsilon neighbourhood you drew. How do we know that those elements outside the epsilon neighbourhood do not ruin the boundedness? Is it because there are only finitely many of them?

[xoppa]

Interesting, for your proof at 3 minutes that the sequence { (-1)^n } diverges, any epsilon less than 1 will produce a similar contradiction. The epsilon value of 1 is the largest of such epsilons that lead to an absurd result (e.g. a false inequality like 2 < 2).

[sarvesh_soni]

i wanted to ask a conjecture i have, I used calculator, and whatever number i start with, If i take the cos of that number infinitely many times, it converges to the same number

like if we define a sequence (An) : An = cos(cos(cos... n times (n)...))
then We have to prove An is convergent!, Is this a well known convergence? Thank you for the quality content!, i will be joining UG course in some months! this will help