i know Im randomly asking but does anyone know of a tool to log back into an instagram account? I was stupid lost the password. I would appreciate any tricks you can give me
1 hour till my final for calc2 , guys this course is heavy af 4 years later: I am process Engineer working @ Neom now and can tell you I use 0 of this course in my work 😂
When your teacher explains 2 hours for one type of convergence test, this video explains 8 of them for only 12mins..... Thank you for uploading this video! Helped a lot !
Dr. Bazett if you were my calc professor I'd have an A. You explained the ratio test as an extended version of the geometric test and concisely explained the difference when to use the direct vs limit comparison tests! Thank you for this!!!!
I wished he explained where he got that 1/25 from on that first geometric series. "I can pull out 2 factors of 5..." isn't quite clear. Just a random 2 factors?
When he goes over the solution to the geometric problem, notice how the 5 has a power of n +1 and not the n - 1 ? He means you need to “lower” that power to the n - 1 for it to nicely match the numerator. So what will bring the n + 1 to the n-1? Remove 5 twice. Aka you remove a 1/25 from the original problem.
Well Im a lecturer at University of Karbala ( Iraq) and I sent this video to my student cause its simple method to realize which test of series you should choose . thanks to you
I needed this video. Deciding on a convergence test has been a nightmare. I have enough time to practice these approaches, before my exam, and I am confident after this video. I grateful
great great great! thank you so much also thank you for your elevated energy in this question, feels so much healthier and happier to be learnign from someone so excited to share this :)
1) Divergence test: lim a(n) =/= 0 ==> series diverges 2) Comparison test (to p-series/geometric series): a(n) a(n) converges 3) Limit comparison test: lim (a(n) / b(n)) = c > 0 b(n) converges ==> a(n) converges 4) p-series & geometric series: 4a) p-series: Σ(1/n)^p converges p > 1 4b) Geometric series: Σa*q^(n-1) = { a/(1-q) if |q| < 1 { diverges if |q| >=1 5) Integral test: Integral from 0 to oo f(x)dx converges ==> Σf(n) converges 6) Alternating series test: Σ(-1)^(n-1) * b(n) if b(n) > 0, decreasing, lim b(n) = 0 then series converges 7) Root test: lim root(n, a(n)) = L (if L is finite, series converges) 8) Ratio test (d'Alembert) (good for factorials & exponents): lim a(n+1)/a(n) = L if L < 1 ==> series converges absolutely if L > 1 ==> series diverges if L = 1 or doesn't exist, use other test
Dr Trefor, I love calculus but conv/diver series was my least favorite. I saw this video some yrs ago and understood the whole thing. I always felt bad for not coming back to thank you🙁. Just logged in to say, THANK YOU SO MUCH! from Scadinavia
This was a very good video in helping folks understand why they should pick or have to use one test over the other. Very deliberate and clear. Excellent work!
Thank you, Dr. Bazett for creating helpful videos. The quality of your production is amazing. I wish to learn more from you. I hope to make videos similar to yours.
Thank you for this really nice video, Dr. Bazett!! Thank you for going to the details of why you did what you did instead of just presenting the solutions!
I think in the divergence test example we can only conclude when the modulus series is convergent as when the modulus series is divergent then we cannot conclude that is because alternating series has alternating +, -. We have the case of conditionally convergent if sigma|un| is divergent and sigma un is convergent. So it is wrong to conclude if the modulus series is divergent
For \Sigma sqrt(n^3+n)/(n^4-n^2) is the following a valid test for divergence / convergence: sqrt(n^3+n)/(n^4-n^2) < n^2/(n^4-n^2) < n^2/(n^2(n^2-1)) < 1/(n^2-1) < 1/n^2, and in the limit 1/n^2 tends to zero.
10:43 The limit as n approaches infinity of 1/sqrt(n) diverges not coverges. The series is a p-series with p = 1/2... a p series is divergent for all p < 1.
Yes you can also say by alternative test summation[ -1^n/sqrt(n)] = summation[(-1^n-1 )* -1 / sqrt(n)] Here bn= -1/sqrt (n) so bn is negative and by alternative test it divergence Sir got confused there. But overall great explanation
Dude, you can't use p series for an alternating series. It is clearly not of p series. If there is (-1)^n in question, then it is alternating series not p series coz p series is of form 1/n^p
you lost me at this part 1:14....... I'm just terrible at skipping obvious steps.. my problem is that i have to go through each step in the solution in order to understand it.. but books and other materials examples do not provide such a thing.. it only expects you to be able to do all that in your head.
3:12 I believe this statement is false. There are possibilities where instead of -1 there is a sin(nπ) or cos(nπ) or something similar in the series, which can be translated into a -1.
Sir at 3:50 The series in the form summation [(-1)^n / sqrt(n)] but not in form of [ (-1)^n-1/sqrt(n)] To apply alternative series test. So you can also say by alternative test summation[ -1^n/sqrt(n)] = summation[(-1^n-1 )* -1 / sqrt(n)] Here bn= -1/sqrt ( n) so bn is negative and by alternative test it divergence Sir got confused there. But overall great explanation
Aaa, wait, I got it, you wanted to leave n-1 in the exponent and so instead you pulled out one 5 so that 5 in the denominator could be to the power of n-1, right?
this guys speaks so enthusiastically about this subject while I'm stressing about my exam lol
Haha, good luck!
@Oakley Christopher Go to hell scammer
@Draven Callan Go to hell bot #2
Same
i know Im randomly asking but does anyone know of a tool to log back into an instagram account?
I was stupid lost the password. I would appreciate any tricks you can give me
I appreciate the enthusiasm. Helps when I'm watching this at midnight for some last-minute cramming. Very helpful, thank you!
GOod luck!
did you pass?
1 hour till my final for calc2 , guys this course is heavy af
4 years later: I am process Engineer working @ Neom now and can tell you I use 0 of this course in my work 😂
Fadel Almobark 4 months later I’m here for the same reason. Due at midnight tonight lol
@Pete My final is in 2 days
1 year later, I’m here too
@@faya943 good luck
i am here before half an hour
When your teacher explains 2 hours for one type of convergence test, this video explains 8 of them for only 12mins..... Thank you for uploading this video! Helped a lot !
Glad it helped!
😂😂
Dr. Bazett if you were my calc professor I'd have an A. You explained the ratio test as an extended version of the geometric test and concisely explained the difference when to use the direct vs limit comparison tests! Thank you for this!!!!
This is one of the most concise and efficient review videos of series/sequences I've come across in my late-night-before-exam cram
fuck yeah "late-night-before-exam cram"
this video is fireeeee my tests in 2 hours this is exactly what i needed
I wished he explained where he got that 1/25 from on that first geometric series. "I can pull out 2 factors of 5..." isn't quite clear. Just a random 2 factors?
When he goes over the solution to the geometric problem, notice how the 5 has a power of n +1 and not the n - 1 ? He means you need to “lower” that power to the n - 1 for it to nicely match the numerator. So what will bring the n + 1 to the n-1? Remove 5 twice. Aka you remove a 1/25 from the original problem.
I want to be as excited about anything in my life as this guy is about convergence testing
Correction: For the series on the bottom right at 0:25, I think you mean n=2 to infinity, not n=1 to infinity. n=1 entails dividing by 0.
Well Im a lecturer at University of Karbala ( Iraq) and I sent this video to my student cause its simple method to realize which test of series you should choose . thanks to you
I needed this video. Deciding on a convergence test has been a nightmare. I have enough time to practice these approaches, before my exam, and I am confident after this video. I grateful
great great great! thank you so much also thank you for your elevated energy in this question, feels so much healthier and happier to be learnign from someone so excited to share this :)
This is seriously one of the best structured math videos Ive seen, so damn useful. Well done!
1) Divergence test:
lim a(n) =/= 0 ==> series diverges
2) Comparison test (to p-series/geometric series):
a(n) a(n) converges
3) Limit comparison test:
lim (a(n) / b(n)) = c > 0
b(n) converges ==> a(n) converges
4) p-series & geometric series:
4a) p-series:
Σ(1/n)^p converges p > 1
4b) Geometric series:
Σa*q^(n-1) =
{ a/(1-q) if |q| < 1
{ diverges if |q| >=1
5) Integral test:
Integral from 0 to oo f(x)dx converges ==> Σf(n) converges
6) Alternating series test:
Σ(-1)^(n-1) * b(n)
if b(n) > 0, decreasing, lim b(n) = 0
then series converges
7) Root test:
lim root(n, a(n)) = L (if L is finite, series converges)
8) Ratio test (d'Alembert) (good for factorials & exponents):
lim a(n+1)/a(n) = L
if L < 1 ==> series converges absolutely
if L > 1 ==> series diverges
if L = 1 or doesn't exist, use other test
I have a series test now and I didn’t understand how to differentiate them til now thank you
Dr Trefor, I love calculus but conv/diver series was my least favorite. I saw this video some yrs ago and understood the whole thing. I always felt bad for not coming back to thank you🙁. Just logged in to say, THANK YOU SO MUCH! from Scadinavia
wow this fellas enthusiasm has me locked in. Thanks!
This was a very good video in helping folks understand why they should pick or have to use one test over the other. Very deliberate and clear. Excellent work!
Yk it's a quality video when it's helping out student even 5 yrs after uploading!
This was so helpful oh my gosh!! I was so stressed about choosing the right test to use during my exams and this really cleared things up!!
Good luck!!
Thank you for this wonderful video which gives a summary of convergence test techniques. I also like the energy I feel when I watch your videos.
Found this right before my final, you're a life saver!
it’s been a while since I’ve done calc, but this video helped me review the topic ty!
Exactly what I was looking for. Please make another video discussing more examples.
I searched Many Many channels, Finally i got the greatest among all🙏🙏🙏🙏
Thank you, Dr. Bazett for creating helpful videos. The quality of your production is amazing. I wish to learn more from you. I hope to make videos similar to yours.
Good luck!
Thanks so much, Dr. Bazett! I can see the patterns to use to differentiate and evaluate these various series.
We love you Dr. Bazett!
favourite math teacher, best graphics 👌
Your voice makes me feel like I can actually learn this course
Thank you for this really nice video, Dr. Bazett!! Thank you for going to the details of why you did what you did instead of just presenting the solutions!
I just wanted to know what different tests are there.
He just served my purpose 😍, without actually knowing it.
I think in the divergence test example we can only conclude when the modulus series is convergent as when the modulus series is divergent then we cannot conclude that is because alternating series has alternating +, -. We have the case of conditionally convergent if sigma|un| is divergent and sigma un is convergent. So it is wrong to conclude if the modulus series is divergent
For \Sigma sqrt(n^3+n)/(n^4-n^2) is the following a valid test for divergence / convergence: sqrt(n^3+n)/(n^4-n^2) < n^2/(n^4-n^2) < n^2/(n^2(n^2-1)) < 1/(n^2-1) < 1/n^2, and in the limit 1/n^2 tends to zero.
Really helpful, sir! Thank you! Have a test coming over this Wednesday, and I have a feeling that it will go well
Best of luck!
This guy speaks very passionitly and its amazing keep up the amazing vids homie!
Very clear and instructive. Thank you
Seeing it from India...really nice ..
Even I want to know from where are u sir..
EXACTLY WHAT I NEEDED !!
this video was a goldmine for me! Thanks a lot
thank you!! it’s sooo hard to find concise videos like this
10:43 The limit as n approaches infinity of 1/sqrt(n) diverges not coverges. The series is a p-series with p = 1/2... a p series is divergent for all p < 1.
Yes you can also say by alternative test summation[ -1^n/sqrt(n)] = summation[(-1^n-1 )* -1 / sqrt(n)]
Here bn= -1/sqrt (n) so bn is negative and by alternative test it divergence
Sir got confused there. But overall great explanation
Dude, you can't use p series for an alternating series. It is clearly not of p series. If there is (-1)^n in question, then it is alternating series not p series coz p series is of form 1/n^p
It's It's shame I only found you the day before my Calc 2 exam. Oh well, better late than never.
you seem so proud of your progress and im over struggling and giving up
where r u now? i hope u survived it well :)
Thanks from the bottom of my heart. You truly are a legend ❤️❤️❤️
Never have I seen a man so excited about alternating series.
thank you, it really helps me prepare for my final exam !!
Best of luck!
Hey doc very nice presentation .Keep it up..From the philippines
Thanks for saving my math... this is really helpful!
Thank you for uploading this video. Helped a lot.
Forgot all about these test. I remember I had fun with them in university :)
Haven't gone to any classes and now I'm cramming for the test today.
I had trouble with determining what to use, so thank you so much for making this video!
Wow, awesome, just wanted this explanation 👏👏
Glad to hear that!
I appreciate your enthusiasm.
Good work dr!!
watching this before my calc 2 final, thanks for helping me cook
Watching 1 day before exams thanks..
This is so helpful thank you!
literally a life saver
great video, super helpful and much clearer than my professor
Im confused how where the n+1 went in the geometric series?
1/25 = 5^-2 and 5^(n+1)/ 5^-2 = 5^(n-1)
if bases are same then powers are added so we can subract one from the power of 5 to take a 5 as common and he did it twice
I am here enjoying your communication skills 🥰
Fabulous work dear
This video saved my life
8:42
If the sequence was powered by n^2, what should we do?
got my ap exam today thank you🙏🏽
Great summary and quick practice problems for the methods!
That was a great video. Love this
great explanation, thank you!!
Thanks my dude!
Truely amazing no words
10:21 ne^(-n^2), Would using root test faster? I really hate using integral test.
thank you sir so much you are the best of them allllll
Awesome experience of studying from your lectures✨.. really very helpful💫
you lost me at this part 1:14....... I'm just terrible at skipping obvious steps.. my problem is that i have to go through each step in the solution in order to understand it.. but books and other materials examples do not provide such a thing.. it only expects you to be able to do all that in your head.
That’s the step I’m lost as well. I wish they would have explained how to factor those numbers…..
Fast and clear. Good job!!
I love you and I love this video - thank you
3:12 I believe this statement is false. There are possibilities where instead of -1 there is a sin(nπ) or cos(nπ) or something similar in the series, which can be translated into a -1.
Oh sure. Sometimes it can be disguised, but it needs to be there in some way is all I meant.
@@DrTrefor Great video Trefor, truly. Keep up the good work.
awesome, cool to have a refresher for all of the tests
10:39 How can this converge if it's a diverging p-series?
I am here with 30 min left for my exam and binge watching anything that can serve as a shortcut 😭😭. Wish me luck
really helpful.Thank you
Thank you !! You're a saver
wow this was super helpful
bro you made cal soooooooo easy and interesting for us..... thanks a lot.......
Sir at 3:50
The series in the form summation [(-1)^n / sqrt(n)] but not in form of [ (-1)^n-1/sqrt(n)]
To apply alternative series test.
So you can also say by alternative test summation[ -1^n/sqrt(n)] = summation[(-1^n-1 )* -1 / sqrt(n)]
Here bn= -1/sqrt ( n) so bn is negative and by alternative test it divergence
Sir got confused there. But overall great explanation
No. It converges. Check again
Really good video i loved it
Just wondering, for 4:44 doesn't failing the alternating series test mean it doesn't approach 0 which means it is divergent?
So. Freaking. Helpful!
thank you my friend!
Can the near-end root test example be replaced with divergence test, since it's (3/2)^n and goes to infinity?
Yup that works too
@@DrTrefor Thank you!!
Isn’t it 1/10 x (2/5)^n here 1:13 ? I am confused
Aaa, wait, I got it, you wanted to leave n-1 in the exponent and so instead you pulled out one 5 so that 5 in the denominator could be to the power of n-1, right?
what is mean by p-test?
How the heck do you get 1/25 for the second one?
Plug in 1 for n and you get 1/25
Does anyone know what song is played during the last part
sir is it always true that limit as n goes to infinity of an is non zero then the series is divergent(an=terms)
Yup. That is called the divergence test.
Great video sir. Keep it up
What if in 3rd problem limit is in numerator???