Monotonic Sequences and Bounded Sequences - Calculus 2
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- Опубликовано: 5 сен 2024
- This calculus 2 video tutorial provides a basic introduction into monotonic sequences and bounded sequences. A monotonic sequence is a sequence that is always increasing or decreasing. You can prove that a sequence is always increasing by showing that the next term is greater than the previous term. This video also discusses bounded sequences. A sequence can be bounded above or have an upper bound it it has a maximum value it can reach. If there is a minimum value in the sequence, then it has a lower bound or it's bounded below. A sequence that is bounded above and bounded below is said to be bounded. A monotonic sequence that is bounded is said to be convergent. This video explains how to determine the convergence and divergence of a sequence by determining if its a monotonic sequence and if its a bounded sequence.
Improper Integrals:
• Improper Integrals - C...
Converging & Diverging Sequences:
• Converging and Divergi...
Monotonic & Bounded Sequences:
• Monotonic Sequences an...
Absolute Value Theorem - Sequences:
• Absolute Value Theorem...
Squeeze Theorem - Sequences:
• Squeeze Theorem For Se...
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Geometric Series & Sequences:
• Geometric Series and G...
Introduction to Series - Convergence:
• Convergence and Diverg...
Divergence Test For Series:
• Divergence Test For Se...
Harmonic Series:
• Harmonic Series
Telescoping Series:
• Telescoping Series
__________________________________
Integral Test For Divergence:
• Calculus 2 - Integral ...
Remainder Estimate - Integral Test:
• Remainder Estimate For...
P-Series:
• P-series
Direct Comparison Test:
• Direct Comparison Test...
Limit Comparison Test:
• Limit Comparison Test
___________________________________
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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 RUclips TV with the Organic Chemistry Tutor.
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Can you do a video about lower, upper, LUB and GL bounds
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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9:48 why do you need to divide it by A_n+1 for this case, but not for other ones I am still confused as to why this is the case?
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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
26:47 isn't this incorrect. by the definition of a monotonic series a_n >= a_n+1 or a_n
I’m also quite confused by this, since I’ve only ever seen that as the definition, when it clearly does not work here
It strictly decreases after the value n = 2.
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1:10 doesn't that make a horizontal line both increasing and decreasing?
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Why don't you make a separate video on complete bolzano-weierstrass theorem
Very clear
Hello,
From my experience it doesn’t appear to be a good strategy to assume that an
You make the assumption based on your observation by plugging in the numbers.
He should have proved it by mathematical induction
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?
Yes
Hello Organic Chemistry Tutor! In your last example, you said that the sequence IS monotonic even though it's increasing for n
It has a monotonic tail, and if a sequence is bounded and has a monotonic tail, it is also convergent.
Its wring he did it wrong
always a helping tutor than you
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
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.
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".
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Can you just take the derivative to show that’s either increasing or decreasing. If positive increasing… if negative dec
What about 1/x where the sequence is bounded and is always decreasing monotonic but is divergent? Doesn’t that break the rule?
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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
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15:59 i couldn't understand what he said and I don't know how he did that. can someone help me please?
1/x^2 is the same as x^-2 so derivative of x^-2 is -2x^-3 which is simply -2/x^3
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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!
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.
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if d(an)/dn is positive can we confirm that a function is monotically increasing?
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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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Why did you add instead of distribute the "2" in the denominator at 5:52
so monotonic functions can have constant sections?
Nice explation keep it up
n-> infinite, Lim An = 0, which doesn't guarantee it converges.
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Is there a sequence that does not bound by up or down?
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18:00 a mistake. It should be n >= -1/2
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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.
Because the sequence is decreasing.
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?
If f(x) is differentiable, then this certainly holds
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