Monotonic Sequences and Bounded Sequences - Calculus 2

[@TheOrganicChemistryTutor]

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Another topic you put to shame my calc 2 teacher at. He puts no effort in teaching and notes are rushed written for someone who is already a master at the topic. Thank you so much

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THANK YOU SO MUCH BUDDY

[@georgesadler7830]

Professor Organic Chemistry Tutor, thank you for explaining Monotonic Sequences and Bounded Sequences in Calculus Two. I also encountered Monotonic and Bounded Sequences in Advanced Calculus; however, I did not understand Monotonic/Bounded Sequences until I watched and analyzed this great video from start to finish. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

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A brilliant video with very clear explanations. Thank you.

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Sir your lecture is so evident that, after listening it I haven't found any doubt.

[@tevinabeysekera6038]

Thanks for explaining so clearly. My professor throws a bunch of complicated logic and formulas only to explain this simple topic.

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Thanks!

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Thank you this was very helpful! at the right pace and the problems you chose were great!

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*smacks the table* THANK YOU! Thank you very much... Finaly!

[@AlfredAceic]

First time understanding this, uh are an outstanding guy thanks

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[@matin-aorosyls.3142]

Just a clarification, at 30:51, you said it is convergent when n starts at 2. But regardless the value of n (whether it is positive or negative), it is no longer convergent, right?

[@rexd1624]

Yes

[@malr19]

Your the best tutor 😭

[@abdallababikir9154]

1:10 doesn't that make a horizontal line both increasing and decreasing?

[@effendi_j.kamal03]

18:16 why do we have n>=1? Like, if we solve the inequality, the value that comes is -1/2, but if we consider only whole numbers, why didn't we put n>=0? @TheOrganicChemistryTutor

[@DirtyStinky]

I know this is a month old but I believe it is due to the fact that Calc 2 will not include negative values for a nth term. So the original problem is a fraction with n as the denominator and therefore we cannot start n from 0 or else it is undefined. So instead we start it from the 1st term n=1 and go up from there.

[@Rajivrocks-Ltd.]

26:47 isn't this incorrect. by the definition of a monotonic series a_n >= a_n+1 or a_n

[@cubix585]

I’m also quite confused by this, since I’ve only ever seen that as the definition, when it clearly does not work here

[@Peter_1986]

It strictly decreases after the value n = 2.

[@champ6277]

You are a legend. Thank you very much

[@berhanuabebe4545]

Can you do a video about lower, upper, LUB and GL bounds

[@elielweerasingha3638]

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Thanks... I learnt many things about this lecture now

[@abouosalah1741]

28:00 I see what u did with that factorial. SMART!

[@erenstine_na7914]

calculus 2 ota yi tu manga 😭😭 out here thinking i dont know what monotonics ate but thanks to him i got

[@Deltax0428]

20:45 that doesn't mean that the sequence converges if the limit found is 0. It's when the function an converges that its limit is 0

[@hilinanegash8520]

Thank you ❤

[@ngaatuband6937]

Thanks so much, it is very helpful

[@Zaptrap101]

15:53 How is this "bounded, this only has a lower bound?

[@Chrisymcmb]

Why did you add instead of distribute the "2" in the denominator at 5:52

[@thomassimancik1559]

Why should the monotonic sequence at 11:45 be greater than or equal to, shouldn't it just be greater than?

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hi (sorry for bad math)

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[@kidusbahiru4369]

On my calculus textbook by James Stewart, it says that a sub n is less than a sub n+1. But you said a sub n is less than or equal to a sub n+1. Your definition contradicts the textbook. But great video overall

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[@frozenfar7231]

15:59 i couldn't understand what he said and I don't know how he did that. can someone help me please?

[@mohanadnada4670]

1/x^2 is the same as x^-2 so derivative of x^-2 is -2x^-3 which is simply -2/x^3

[@sheesh602]

my appreciation for you do be increasin'

[@MarcheseMethods]

Hello, From my experience it doesn’t appear to be a good strategy to assume that an

[@bottomtext251]

You make the assumption based on your observation by plugging in the numbers.

[@markdonnelviernes3369]

He should have proved it by mathematical induction

[@AtomicSmalls]

I thought, because of the growth rate theorem, n! infinity 3^n/n! would be an infinite series. Thus would make it a monotonic, lower bounded, divergent series

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exams tomorrow! helped me quite a bit :D

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Quality content 😍

[@sana_136]

Very helpful 👍🏻

[@ryeonnaderi6731]

9:50 why are we dividing an/an+1, because on the first problem we did not have to divide so I am confused on that.

[@frankanonymous3766]

Because the sequence is decreasing.

[@hellohi-mj8ho]

What about 1/x where the sequence is bounded and is always decreasing monotonic but is divergent? Doesn’t that break the rule?

[@MuhamamdUmar-lp4ee]

I have a question What if the sequence is bounded above but it is decreasing function Still it is monotonic and also bounded but will it be convergent?

[@LongArmsShortTorse]

I find the concepts of sequences bounded above and/or bounded below kind of hard to understand because the semantic is kind of counter-intuitive for instance bounded below means that the sequence has a lower bound, but from face value it sounds like the sequence would be below a limit by the words "bounded below".

[@mwnhasarnarzary9694]

Why don't you make a separate video on complete bolzano-weierstrass theorem

[@geennaam7129]

around 28:30 shouldn't it be n-1?

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[@abd......]

Sir very well done.. Great explanation

[@ИванПетрович-г6ю]

18:00 a mistake. It should be n >= -1/2

[@adamjahani4494]

Can you just take the derivative to show that’s either increasing or decreasing. If positive increasing… if negative dec

[@AwuraamaDadson]

Thank you again ❤️❤️

[@leenali8741]

The graph at minute 14:03 is not monotonic, is it?

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Albert Einstein How to do math?? Step 1= write the question. Step 2= cry....

[@youtubeproject4831]

if d(an)/dn is positive can we confirm that a function is monotically increasing?

[@nicholasmaloof8378]

17:57 I feel dumb, but what happened to the denominator? Why is it just (n+1)^2 >= n^2?

[@lightx8126]

Cross multiplication

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Is there a tip/trick for me to keep up with the math needed to understand this better? I find myself understanding things when he specifically explains them (in detail) but If I was supposed to intuitively get it right I just wouldn't have it on my mind. Thanks!

[@louaih]

If you're talking about the algebraic tricks he used, such as manipulating inequalities, I couldn't hope to explain it better than @TheOrganicChemistryTutor. However, my main takeaway from the entire video is that a sequence converges when it is bounded and monotonic. A bounded sequence is a sequence that has a floor and ceiling (lower and upper bounds) as n goes to infinity. A monotonic sequence is a sequence that only increases or only decreases in the long run. Note how in the last example, the sequence wasn't monotonic on n >= 1, but it was monotonic on n >= 2.

[@shantanrs2827]

so monotonic functions can have constant sections?

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Your a legend

[@kelvinndereba5423]

Thankyou for the video

[@butx1555]

22:00 important theorem

[@Zaptrap101]

This video is so confusing, 30:41 if so then doesn't it disobey the rule of monotonic that a1 a1 >= a2 >= a3 ...

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God bless you

[@amagiyaisaiah692]

Very clear

[@suicider8062]

always a helping tutor than you

[@halfdonut6505]

Hello Organic Chemistry Tutor! In your last example, you said that the sequence IS monotonic even though it's increasing for n

[@benclarke7263]

It has a monotonic tail, and if a sequence is bounded and has a monotonic tail, it is also convergent.

[@ruger51995]

Its wring he did it wrong

[@phandinhthanh2295]

I have a question: is it true that whenever we want to determine the monotonicity of a sequence {a_n} we could always use the Calculus method of taking derivative of continuous function f(x) with f(n) = a_n? And if f(x) is increasing on an interval then {a_n} is increasing on that interval and vice versa if f(x) is decreasing on an interval then {a_n} is decreasing on that interval? Is this obvious or is there some theorem/corollary stating this?

[@kwerty4761]

If f(x) is differentiable, then this certainly holds

[@subashkafle]

can someone verify what he said at 19:16 bounded above 1 or bounded below 1 similarly for 0 as well. is it bounded above 0 or below 0. I think he just mixed up but may be I am wrong.

[@nicholasnyongesa9673]

bounded below at 0 and bounded above at 1

[@bottomtext251]

What he said means: upper bound at 1 and lower bound at 0

[@Zaptrap101]

​@@nicholasnyongesa9673why is bounded above at 1?

[@Zaptrap101]

​@@bottomtext25113:20 It is the exact image here, but why is this case only bounded below but not above?

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