@@onwabadosini2962 indeed the limit as n approaches infinity of 1/3^n =0 but in the 3 problem we are limit comparing (1/3^n +5 * 3^n /1) = lim(3^n / 3^n + 5) and that equals 1
Bro you are really making the difference in this world. Thank you for helping us.. Tomorrow's Engineers and other professionals would always keep you in ours minds, because you just made it easy for us to closed the open gaps in the concept.
0:22-1:22 For anyone who is confused about why the last question is divergent when the limit=1, this explains it. This is an important rule to remember, it is simply a proof. The limit doesn't determine the answer. It just proves that the answer found for one series (bn in this case) is going to be the same for the other (an).
You may actually be the reason i pass my engineering maths test this coming Monday. Really helpful explanation video. I had not yet come across such clear and concise explanation. Thank you
You are doing wonders for Engineers during quarantine 💖 Many universities are closed and it is challenging to navigate course work and keep up from home. You videos really make a difference :)
Professor Organic Chemistry Tutor, thank you for another great video/lecture on the Limit Comparison Test in Calculus Two. This is a very important test in determining convergence and divergence of a series. Learning different tests for convergence/divergence can be a challenge for all students in Calculus. Thís is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
Thank you again for the 56th time for helping me understand and review my calculus, physics and organic chemistry. If I pass my semester I owe it to you.
@@Daniilope Don't give up! Calculus can be difficult to understand. If you understand the reason behind using a certain formula or type of limit you'll find it easier to remember. It's a lot of practice. Wish you the best!
In the second example, you have to calculate the first term of the series, n=0, therefore obtaining the series that starts at n=1. After that, you just have do exactly what JG did about the harmonic series. When you see that the new series is divergent and the limit test gives you 1, you can unify the first term to the series that starts at n=1, and then obtain the one the you had a the beginning.
Awesome video!!! I love the way you teach us like you have a God-given gift. God bless you you are a grade saver! I literally don’t even care that I don’t get the material in the class cuz In the end you are my real teacher always! ✍🏻❤️
taking 4 summer classes. have my second calc exam next Wednesday. if I pass this exam, actually no. if I pass this semester in calc 2, I owe it all to you. calc 2 in 10 weeks is a big beast but im currently kicking its ass
2:33 we now finally know what Gunna and Future were talking about when they said they were pushing P, they were talking about Calc 2 Limit Comparasion test, p=3. They're ahead of the rap game so much man, Drake's out there sampling Alvin and the Chimpunks and they're out here sampling Limit Comparasion Tests and Differential Calculus. Kudos to them and OCT! 🤲🤲
for those who didnt understand the part in #2 where he removed the 5 in the limit, you can use L'Hopital on the limit which is ln3*3^n/ln3*3^n which cancels out to 1
Because (an) is a power series, its power is the root( √) which is less than 1, since the rule for power series convergence states that P must be >1,(thomas calculus) book-11th edition on page 792. (an) here is divergent.
@@ludovico-291hzu He's basically multiplying it by one but in a different form. 1/n divided by 1/n is 1. The denominator of sqrt(1/n^2) actually simplifies to 1/n. The reason that he puts it under the square root though, is so he can multiply the radicands and just keep it under the square root.
I'm still a bit confused. If the limit comparison test depends on the regular comparison test to tell if its convergent/divergent, what is the point of the limit comparison test? Isn't the limit comparison test just an unnecessary extra step?
n can not equal zero as your bn in the second example, in order to correct this i believe you should set both starting at n=1 and then add one to the n in the an.
If you watched the direct comparison test video he has, hes subtly nudging that test when he goes to prove b_n or an as either convergent or divergent.
Thank you for this video! Could you please do another video like this but working with e?? Most of my homework and practice quizzes have those types problems.
This only applies when the comparison test rule is flipped -> comparison test rule: bn converges then an converges and an divergent then bn dgt. It can’t be the other way around therefore you apply limit comparison test
What happens if you have a constant in both the numerator and denominator. 1. Which do you get rid off, if not both. 2.if the constant remains in either num or denim, can we disregard it when dividing an and bn?
I don't understand the second example, where the first term is already divergent. Shouldn't the series start with n=1? In this case there was no problem but when we use 1/n^2 for comparison, (1/sqrt(n^4+1) on the left)we will get different answers(convergent if starting from n=1, divergent if the series starts with n=0) which is probably not the case in the series we want to test. What is wrong with this situation?
couldnt you use the direct comparison test for the last example? Because 1/(3^n + 5) is less than 1/3^n and 1/3^n converges, so by the direct comparison test, 1/(3^n + 5) would converge?
Isn't the limit comparison test flawed for some examples? If we were to have the sum from n=1 to infinity of 1/(n^2-1) and tried to evaluate whether it converges or diverges, the closest thing to compare it with would be the sum from n=1 to infinity of 1/(n^2). Using the limit comparison test, we would see that they also reach some value L, so that would mean that since (1/(n^2)) is convergent, 1/(n^2-1) should also be convergent. However, the first value of the sum in 1/(n^2-1) is 1/0, which should make it impossible for them both to converge. Are there some rules that I have missed?
for the second example, why are you able to use 1/n when if sub n=0 , it is not finite which violates one of the constraints which says that the expression must be positive an finite for all values for n
I know the 5 is insignificant in 3^n / 3^n + 5 but that wouldn't fly with my professor (who is super into doing proofs and having us do proofs). The actual method would be to multiply the top and bottom by 1/3^n to get the form 1/(1+5/3^n) which is provably 1
If the bn (big series) is divergent, you cannot say that an is divergent as well or if an is convergent, you cannot say bn is convergent. Therefore you need to use limit test. But there was a few examples in this video can solve with direct comp. test too but I guess the instructor missed that... In short, you need use the limit test when you can't use the comparison test. You can also use the limit test at first but it will be very tedious.
FAIRA Fa yeah, that’s true. Thank you for your help, I was just kinda skeptical whether or not it is “legal” to go from 1/3^n to (1/3)^n. But your explanation made it clear, so thanks again☺️
8:12 I'm a little confused. I thought bother series would converge since you got a finite sum with the limit comparison test of lim An/Bn, so if An or Bn diverge by themselves, then they bother diverge even after the LCT?
@@saif0316 If your answer is: > 0 , both series either converge or diverge =0 and Bn converges, then An also converges =Infinity and Bn diverges, then An also diverges
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Think you made a mistake in the third one, isn't the limit as n approaches infinity of 1/3^n equal to zero ? Please confirm.
They were talking about sum of 1/3/^n@@onwabadosini2962
@@onwabadosini2962 indeed the limit as n approaches infinity of 1/3^n =0 but in the 3 problem we are limit comparing (1/3^n +5 * 3^n /1) = lim(3^n / 3^n + 5) and that equals 1
Bro you are really making the difference in this world. Thank you for helping us.. Tomorrow's Engineers and other professionals would always keep you in ours minds, because you just made it easy for us to closed the open gaps in the concept.
BROOO facts, I’m a mechanical engineer and I can say personally you’re helping the world genuinely🙏🏻🙏🏻
Fact man, I study petroleum engineering, this guy helps me a lot since freshman❤️
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I crushed calculus 2, thanks to this channel, and now I'm in graduate school enjoying these as a quick reviews of the tedious mechanics of calculus
I recommend watching the direct test first and then the limit one because it’s easier to understand
100% true
u are 100% correct
0:22-1:22 For anyone who is confused about why the last question is divergent when the limit=1, this explains it. This is an important rule to remember, it is simply a proof. The limit doesn't determine the answer. It just proves that the answer found for one series (bn in this case) is going to be the same for the other (an).
Thank you!!
thanks! couldn't have said it better
You mean the one before last?
You may actually be the reason i pass my engineering maths test this coming Monday. Really helpful explanation video. I had not yet come across such clear and concise explanation. Thank you
Whenever I have confusion, I search your video and understand what going on. Thanks bro!
You are doing wonders for Engineers during quarantine 💖 Many universities are closed and it is challenging to navigate course work and keep up from home. You videos really make a difference :)
I have a test on this Monday and I plan to binge watch these videos and make a much needed A on this test!! Thank you sooo sooo much!!
fail
Last minute studying isn't gonna work. You should've studied earlier. I should take my own advice too. My test us tomorrow. So yay
Organic chemistry is a life saver
@Tshilady Tshilady you really took all the time in the world to respond to some fucker from 2 years ago and said something mean???
@@Jackielikagovernor ur the mean dude here lmao
Professor Organic Chemistry Tutor, thank you for another great video/lecture on the Limit Comparison Test in Calculus Two. This is a very important test in determining convergence and divergence of a series. Learning different tests for convergence/divergence can be a challenge for all students in Calculus. Thís is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
I love you, thanks for the videos. I wouldn't be passing calc 2 otherwise- your videos are well explained and concise!
I get so happy every time a see you made a video on a subject I need to study 🙏
Thank you again for the 56th time for helping me understand and review my calculus, physics and organic chemistry. If I pass my semester I owe it to you.
Did you pass the semester
@@Daniilope I did 😁 I almost didn't pass organic chemistry but I made it. This channel helped me pass Calculus 1, Calculus 2 and physics.
@@arianeparadis6439 that’s so good to hear! Currently calculus is kicking my butt so hopefully this channel works it’s magic like it did on you 😭🙏
@@Daniilope Don't give up! Calculus can be difficult to understand. If you understand the reason behind using a certain formula or type of limit you'll find it easier to remember. It's a lot of practice. Wish you the best!
how do we know if we should use the limit comparison test or the direct comparison test?
If one doesn’t work, then use the other
@@Cobenski4 ty
The way you teach is INSANE ! 😍
I have my calculus 2 final in 1 hour. Sir, you just cleared my doubts about the limit comparison test. Thank you for so much.
I'm curious too...can't just leave us hanging dude
I guess you didn't learn the limit comparison test as well as you had hoped.
How did you do?!?
Well, I graduated college 2 months ago but I did pass the test. I remembered that I got an 87 on that final and now I work in tech. Success story 😂
@@kelvincheng999 you give me hope, I have a calc 2 midterm tomorrow and hopefully will pass.
In the second example, you have to calculate the first term of the series, n=0, therefore obtaining the series that starts at n=1. After that, you just have do exactly what JG did about the harmonic series. When you see that the new series is divergent and the limit test gives you 1, you can unify the first term to the series that starts at n=1, and then obtain the one the you had a the beginning.
Amazing video, literally running through my math problems because of your extraordinary explanations! Thank you!
You're the OG homie. Thank you once AGAIN for enlightening us peasants about another math concept. Keep rocking man!
Awesome video!!! I love the way you teach us like you have a God-given gift. God bless you you are a grade saver! I literally don’t even care that I don’t get the material in the class cuz In the end you are my real teacher always! ✍🏻❤️
your voice is so soothing .i'm so happy i found your channel
Man has been saving me the past 7 years🙏😭
You are the best professor I have ever seen .
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I love you, never gonna stop loving you, youre getting me through calculus and into an engineering degree.
Thanks my guy, I usually watch khan, but this filled a lot of gaps in my understanding that i couldn't get with khan.
You’re AMAZING you make everything so much easier i loved it! Thank you so much
taking 4 summer classes. have my second calc exam next Wednesday. if I pass this exam, actually no. if I pass this semester in calc 2, I owe it all to you. calc 2 in 10 weeks is a big beast but im currently kicking its ass
You are saving my university grades for Calc II. I love you!
wouldn't the example at 8:17 be solvable through the direct comparison test?
So would the first example.
Got my final in 6 minutes and ofc this guy saves the day again haha
last problem can be solved using direct comparison test
really appreciate your effort man, thanks.
When you plot the series, both converge to the cero. 1/n as n is greater and greater the sum is approaching to cero (0)
Thanks! I was looking for how to solve one of these problems algebraically and what you did made perfect sense.
2:33 we now finally know what Gunna and Future were talking about when they said they were pushing P, they were talking about Calc 2 Limit Comparasion test, p=3. They're ahead of the rap game so much man, Drake's out there sampling Alvin and the Chimpunks and they're out here sampling Limit Comparasion Tests and Differential Calculus. Kudos to them and OCT! 🤲🤲
Thank you so much for this, you don't know how much this helped me... all ur vids have basically taught me more than my teachers at school lol
for those who didnt understand the part in #2 where he removed the 5 in the limit, you can use L'Hopital on the limit which is ln3*3^n/ln3*3^n which cancels out to 1
thank you!
Watching you from Brazil, you're the best
just a heads up, you have the harmonic series at 5:49 going from 0 to inf.. but its undefined at zero
Dude thanks again. I was struggling with this, but thanks to you I’m back on track. 🐐
Youre the reason I'm passing with A's on exams in Calc 2
imagine getting A's in calc 2..i wish.
@@megsperry2925 fr lmaoooo, only in my dreams
@@adansilva6851 I'm only just trying to pass..with a nice 'ol C
@@megsperry2925 im aiming for a high grade but as long as I pass, i'll take it
@@adansilva6851 good luck :, )
WHAT KIND OF LEGEND ARE YOU
Thanks again OCT. I can now sleep soundly in my bed knowing how to execute the LCT. You are the best.
If Bn diverges and if u take the lim and get 1 which is a finite number then why does An diverge? Ref. 8:10
Same doubt
Because (an) is a power series, its power is the root( √) which is less than 1, since the rule for power series convergence states that P must be >1,(thomas calculus) book-11th edition on page 792. (an) here is divergent.
You are my math god
if i can get through my physics program, i will be proud to say that youre basically my instructor 😊
sa makakita ani nga cmate, hiiii
Can anyone please explain the step at 6:55? When he multiplied the top and bottom by 1/n. Thanks!
if you understand that, can you please explain it to me?
@@ludovico-291hzu He's basically multiplying it by one but in a different form. 1/n divided by 1/n is 1. The denominator of sqrt(1/n^2) actually simplifies to 1/n. The reason that he puts it under the square root though, is so he can multiply the radicands and just keep it under the square root.
Thank you, but I think the summation in the series b_n should starts from 1 not from zero.
I'm still a bit confused. If the limit comparison test depends on the regular comparison test to tell if its convergent/divergent, what is the point of the limit comparison test? Isn't the limit comparison test just an unnecessary extra step?
How can you teach soooo efficiently
Thank u sooo much sir for ur kindness and for helping us
i just became your fan ....after watching "comparison tests videos".....what an explanation👌👌....love you Sir 🙏
👉love from India ✌️👌
Thanks for the video! You are an awesome teacher!
n can not equal zero as your bn in the second example, in order to correct this i believe you should set both starting at n=1 and then add one to the n in the an.
Best teacher everrrrr
You are a life saver, Thank you !
يسطا لو فهمتها ياريت تشرحهالي عليا اختبار بعد بكرا ده الانستا بتاعي . (_5stc) ولو ملقتهوش خد ده endl9
@@moazosama معلش والله مخدتش بالي من الريبلاي إلا دلوقتي سوري😂. أتمنى تكون أديت كويس.
excelente video, muy practico para el aprendimiento rapido.
On example 2 I’m confused how you chose to divide by 1/n and sqrt of 1/n^2
Thanks you so much for making these videos! They really help!
I'm just wondering, but isn't the last problem technically a Direct Comparison Test?
@@mundheral-ahmadi8939 I was thinking the same thing. I think he's just taking it one step further.
If you watched the direct comparison test video he has, hes subtly nudging that test when he goes to prove b_n or an as either convergent or divergent.
Your videos are immensely helpful and have gotten be through Organic Chemistry and now Calculus II. I would have failed without you so thank you!
What did you make in orgo? i take it next semester
Thank you for this video! Could you please do another video like this but working with e?? Most of my homework and practice quizzes have those types problems.
Aww! I literally just had my test 3 hours ago! It’s okay tho, I still love you ♥️
Hope you scored high!
you sir have blessed me
Excelent explanation sir
Its good and all but when we have the comparison test, why do we need this?
This only applies when the comparison test rule is flipped -> comparison test rule: bn converges then an converges and an divergent then bn dgt. It can’t be the other way around therefore you apply limit comparison test
What is the some real-life application of comparison test?with examples.
What happens if you have a constant in both the numerator and denominator.
1. Which do you get rid off, if not both.
2.if the constant remains in either num or denim, can we disregard it when dividing an and bn?
Joseph George pull the constant out in front of the limit
Joseph George use limit laws
I’ve been watching this dude for years now 💀
Literally carrying me 😂
I don't understand the second example, where the first term is already divergent. Shouldn't the series start with n=1?
In this case there was no problem but when we use 1/n^2 for comparison, (1/sqrt(n^4+1) on the left)we will get different answers(convergent if starting from n=1, divergent if the series starts with n=0)
which is probably not the case in the series we want to test. What is wrong with this situation?
couldnt you use the direct comparison test for the last example? Because 1/(3^n + 5) is less than 1/3^n and 1/3^n converges, so by the direct comparison test, 1/(3^n + 5) would converge?
bingo
Bravo, maestro
Sir how to take bn in limit comparison test
Why do you multiply the numerator and denominator with 1/n at 6:55?
carrying my degree rn
How do we know which one is supposed to be a sub n and which one should be b sub n?
Thank you sooooo much
Isn't the limit comparison test flawed for some examples? If we were to have the sum from n=1 to infinity of 1/(n^2-1) and tried to evaluate whether it converges or diverges, the closest thing to compare it with would be the sum from n=1 to infinity of 1/(n^2). Using the limit comparison test, we would see that they also reach some value L, so that would mean that since (1/(n^2)) is convergent, 1/(n^2-1) should also be convergent. However, the first value of the sum in 1/(n^2-1) is 1/0, which should make it impossible for them both to converge. Are there some rules that I have missed?
4:47 not to mention 1/n^3 is bigger than a of n so DCT would work too
The Organic Chemistry Tutor you are too much
Thank you so much for this
for the second example, why are you able to use 1/n when if sub n=0 , it is not finite which violates one of the constraints which says that the expression must be positive an finite for all values for n
Yes I was thinking about that too, maybe you should specifies in parenthesis for an n =/= to 0
just think math easy after you men ....too much respect for your work lol
@7:00 there was a much easier way of doing all that but yeah
I know the 5 is insignificant in 3^n / 3^n + 5 but that wouldn't fly with my professor (who is super into doing proofs and having us do proofs). The actual method would be to multiply the top and bottom by 1/3^n to get the form 1/(1+5/3^n) which is provably 1
That's dumb.
U help me a lot thanks for that
what if the L is 0? I know it is finite but is it positive?
What happens when the limit of An/Bn is infinity?
the limit comparasion test is inconclusive (not 100% tho)
Why can't you use the normal comparison test for the first one?
so are you saying that if you get a finite number for the limit comparison then both are the same as the results you got from the direct comparison"?
What is the difference between The Direct Comparison Test and The Limit Comparison Test?
If the bn (big series) is divergent, you cannot say that an is divergent as well or if an is convergent, you cannot say bn is convergent. Therefore you need to use limit test. But there was a few examples in this video can solve with direct comp. test too but I guess the instructor missed that... In short, you need use the limit test when you can't use the comparison test. You can also use the limit test at first but it will be very tedious.
Thank you, sir.
FAIRA Fa Hey, @8:53, do you know how 1/3^n became (1/3)^n? Thanks
@@uzumakikuroko9000 yes because 1^2=1, 1^(2000)= 1
1 exponent anything is one.
So you still have the same term 1/(3)^n
I hope it helped.
FAIRA Fa yeah, that’s true. Thank you for your help, I was just kinda skeptical whether or not it is “legal” to go from 1/3^n to (1/3)^n. But your explanation made it clear, so thanks again☺️
You are awesome, sir. Keep it up!
so, for the limit comparison test you do not need to compare if the new series b_n is bigger or smaller than a_n?
Wish I could like the video 100000x
8:08 the harmonic series diverges and the original series has a finite limit which is 1 and thus converges. Clear up the misconception
8:12 I'm a little confused. I thought bother series would converge since you got a finite sum with the limit comparison test of lim An/Bn, so if An or Bn diverge by themselves, then they bother diverge even after the LCT?
Ryan if you know Bn diverges and you do the test and get a finite number then An also diverges.
6 months later I still wonder
Parker Mell and if you dont get a finite number?
@@saif0316 If your answer is:
> 0 , both series either converge or diverge
=0 and Bn converges, then An also converges
=Infinity and Bn diverges, then An also diverges
You are so helpful!
you are a life saver
💌
For the first example, since is bn is a bigger series and converges couldn't you say the an series converges by regular comparison test?
I guess