Final Exams and Video Playlists: www.video-tutor.net/ Full-Length Videos & Worksheets: www.patreon.com/MathScienceTutor/collections Next Video: th-cam.com/video/tHy3TXmZpF0/w-d-xo.html
I remember watching your videos before the January 2020 algebra regents exams and they have helped me a ton. You are one of the main reasons why I passed. My teachers would never explain things to me the way you do. You make everything seem like a piece of cake compared to my teachers. Just wanted to give you a big thanks.
Followed u for a long time now, thank you for helping my calculus, my Ap physics, my chemistry 12. And now even helping me when I am in uni now. You are a true lad, and deserve the sub. Thank you again for helping me get into Imperial college.
You're amazing and I love you. Thank you so much for everything you do!!!!! I have used you for Organic Chemistry, Calculus 1, Calculus 2, Physics 1, and Physics 2.
Professor Organic Chemistry Tutor, thank you for using multiple well-known examples to explain/analyze Converging and Diverging Sequences in Calculus Two. In all Calculus textbooks, there is a whole chapter on Sequences and Series. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
@@FerghusCameron Just to take the n to an easier place. This can be solved by many ways, and you can keep the n and evaluate it to find the answer (if it's converging or diverging) but dividing by 1/n takes the n to 5/n which leads to the final answer easily (because of the rule 1/n=0).
@@FerghusCameronif you took the limit as n approaches infinity, you’d end up with infinity over infinity. By multiplying by the numerator and denominator by 1/n, you are able solve. Another method of solving for the infinity/infinity problem is applying L’Hopital’s Rule, which would also give you the same answer and is more applicable in vague situations where multiplying by 1/n isn’t available or intuitive.
I have been stuck on Sequences and Series for three days now. I don’t understand it, but your videos cleared it up a bit. I really hope the next bit of your videos help because I don’t have any other ideas.
for anyone else scratching there head at this! plugging in inf into (1+1/n)^n = (1+1/inf)^inf = (1+0)^inf = (1)^inf = 1^inf = undefined this is why we had to solve it
why would you add the last step at 22:15? 1/(n+2) is just 1/infinity which is zero. Absolutely no need to multiply top and bottom to get the zero in the numerator
why do u use l'hopital rule insted of just substitute infinity in the equation lim n-> infinity n * sin(1/n)??? and all your video are just unreal. u made life easy for us.
It was just to avoid L'hopital's rule. It doesn't change the limit if you multiply the top and bottom of the fraction by any real number other than zero, infinity and negative infinity.
At 13:49, why can you determine it's convergent since lim is 0 here? The divergence test on our slide said that you can't determine whether it's convergent or divergent straight forward if lim=0. I'm so confused:(((
I think you're getting confused between the convergence and divergence of series and of sequences, the divergence test of series needs the limit you're talking about, and in this example we can't determine whether the infinite sum is convergent or divergent(the series) but what we can determine in this example is that the sequence will end up at 0 when we approach infinity
I'm lost at 19:08. Apparently the answer is now 1, but why couldn't it be 0? sin(1/n) would return 0 and anything multiplied by 0 is 0. Why is that invalid? Are we only able to find the sequence when n only appears once in the function?
Never mind i understand now. The sin of 1/n may approach zero but it never reaches zero. Then it is multiplied by what ever n is as it approaches infinity.
29:20 Am I missing something? The n-th term test specifically states that if the limit is a real, non-zero number, the infinite sum of the series diverges.
@25:35 I felt there was an easier solution, maybe I am mistaken? It's been a long time since I've taken calc but someone please correct me if I'm wrong! (1+1/n)^n The limit as n --> infinity for 1/n is 0 So (1+0)^n 1^n, as n--> infinity, 1 to any power is 1... so the limit is 1.
1^inf is an indeterminate form because the 1 at the base is not an exact 1, it is a limit, so it is a number that approaches 1. Therefore, it cannot be the considered the same as a power of 1. however, i will be soon posting on my channel a video on indeterminate forms of limits, you could subscribe if you wish to follow it up.
because you want to eliminate the n on the top and bottom. Whatever you do to the top, you do to the bottom. Therefore, you can multiply the top and bottom by 1/n
Final Exams and Video Playlists: www.video-tutor.net/
Full-Length Videos & Worksheets: www.patreon.com/MathScienceTutor/collections
Next Video: th-cam.com/video/tHy3TXmZpF0/w-d-xo.html
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I remember watching your videos before the January 2020 algebra regents exams and they have helped me a ton. You are one of the main reasons why I passed. My teachers would never explain things to me the way you do. You make everything seem like a piece of cake compared to my teachers. Just wanted to give you a big thanks.
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Followed u for a long time now, thank you for helping my calculus, my Ap physics, my chemistry 12. And now even helping me when I am in uni now. You are a true lad, and deserve the sub. Thank you again for helping me get into Imperial college.
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Nursultan Beloved this is only the first section in the first chapter in my course and at most only one question will come on it in the test.
@@theadel8591 I feel like that'll be the case for us, except I bet most of it is going to be power series
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Professor Organic Chemistry Tutor, thank you for using multiple well-known examples to explain/analyze Converging and Diverging Sequences in Calculus Two. In all Calculus textbooks, there is a whole chapter on Sequences and Series. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
love your style of teaching and your voice, thank you for saving my life
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At 4:30 why do you multiply the top and bottom by 1/n? How do I recognize when to use this?
Did you find out the reason? I just started Pre calculus and would like to know thr proof for multiplying the top and bottom by 1/2
@@FerghusCameron
Just to take the n to an easier place. This can be solved by many ways, and you can keep the n and evaluate it to find the answer (if it's converging or diverging) but dividing by 1/n takes the n to 5/n which leads to the final answer easily (because of the rule 1/n=0).
@@FerghusCameronif you took the limit as n approaches infinity, you’d end up with infinity over infinity. By multiplying by the numerator and denominator by 1/n, you are able solve. Another method of solving for the infinity/infinity problem is applying L’Hopital’s Rule, which would also give you the same answer and is more applicable in vague situations where multiplying by 1/n isn’t available or intuitive.
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I have been stuck on Sequences and Series for three days now. I don’t understand it, but your videos cleared it up a bit. I really hope the next bit of your videos help because I don’t have any other ideas.
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Yes it is cause the lim as n>infinity of ln(1+(1/n))/(1/n) is ln(1+0)/(0) which is undefined.
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for anyone else scratching there head at this!
plugging in inf into (1+1/n)^n = (1+1/inf)^inf = (1+0)^inf = (1)^inf = 1^inf = undefined
this is why we had to solve it
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why would you add the last step at 22:15? 1/(n+2) is just 1/infinity which is zero. Absolutely no need to multiply top and bottom to get the zero in the numerator
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why do u use l'hopital rule insted of just substitute infinity in the equation lim n-> infinity n * sin(1/n)???
and all your video are just unreal. u made life easy for us.
This guy is supposed to do all math-related videos. He's genius
Wait so at 22:55 why do you multiply by 1/n instead of just evaluating the limit there??
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Will give u a good idea about convergence nd divergence.
But formaly the idea of Convergence nd divergence is different.
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for the question at 13:44 couldn't you just simplify sin(n)/n to 1? Is that an identity that can be used?
At 5:10 why was the 1/n adjustment done?
It was just to avoid L'hopital's rule. It doesn't change the limit if you multiply the top and bottom of the fraction by any real number other than zero, infinity and negative infinity.
thanks i love ur videos so much
for 13:21 why can’t we use lopital rule and make it lim cos(n) = 1?
7:10 even though it seems counter-intuitive it is technically a continuous function given the ε-δ definition of continuous functions.
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Could you not have used L'Hopital's rule for the limit as n approaches infinity of a_n = 8n/3n-5? (5:35)
Your videos are great! Thank you. I wonder if someone could help me understand why there is a vertical asymptote at 5/3? Confused!!!
At 13:49, why can you determine it's convergent since lim is 0 here? The divergence test on our slide said that you can't determine whether it's convergent or divergent straight forward if lim=0. I'm so confused:(((
I think you're getting confused between the convergence and divergence of series and of sequences, the divergence test of series needs the limit you're talking about, and in this example we can't determine whether the infinite sum is convergent or divergent(the series) but what we can determine in this example is that the sequence will end up at 0 when we approach infinity
I'm lost at 19:08. Apparently the answer is now 1, but why couldn't it be 0? sin(1/n) would return 0 and anything multiplied by 0 is 0. Why is that invalid? Are we only able to find the sequence when n only appears once in the function?
Never mind i understand now. The sin of 1/n may approach zero but it never reaches zero. Then it is multiplied by what ever n is as it approaches infinity.
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Watching this 20 minutes before my final
Should not we apply stolz-cesara instead of l’hospital when dealing with sequences? Please explain!
29:20 Am I missing something? The n-th term test specifically states that if the limit is a real, non-zero number, the infinite sum of the series diverges.
I think you're thinking about series instead of sequences.
Well done hussy!
@25:35 I felt there was an easier solution, maybe I am mistaken? It's been a long time since I've taken calc but someone please correct me if I'm wrong!
(1+1/n)^n
The limit as n --> infinity for 1/n is 0
So (1+0)^n
1^n, as n--> infinity, 1 to any power is 1... so the limit is 1.
1^inf is an indeterminate form because the 1 at the base is not an exact 1, it is a limit, so it is a number that approaches 1. Therefore, it cannot be the considered the same as a power of 1. however, i will be soon posting on my channel a video on indeterminate forms of limits, you could subscribe if you wish to follow it up.
at 4:48 why do you multiply by 1/n on the numerator and denom?
because you want to eliminate the n on the top and bottom. Whatever you do to the top, you do to the bottom. Therefore, you can multiply the top and bottom by 1/n