Choosing Which Convergence Test to Apply to 8 Series
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- Опубліковано 19 лис 2024
- Deciding which convergence test to apply to a given series is often the hardest part of the unit on series convergence. In this video, I'm going to loosely walk through some larger strategies for picking and choosing the method, focusing on 8 different series and 8 different methods:
Geometric Series
Integral Test
Divergence Test
Alternating Series Test
Comparison Test
Limit Comparison Test
Ratio Test
Root Test
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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
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
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?
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"
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
this video is fireeeee my tests in 2 hours this is exactly what i needed
I want to be as excited about anything in my life as this guy is about convergence testing
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
I have a series test now and I didn’t understand how to differentiate them til now thank you
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 :)
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.
Yk it's a quality video when it's helping out student even 5 yrs after uploading!
This is seriously one of the best structured math videos Ive seen, so damn useful. Well done!
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!
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
I searched Many Many channels, Finally i got the greatest among all🙏🙏🙏🙏
Your voice makes me feel like I can actually learn this course
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!!
I just wanted to know what different tests are there.
He just served my purpose 😍, without actually knowing it.
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!
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.
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
favourite math teacher, best graphics 👌
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.
We love you Dr. Bazett!
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.
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!
Found this right before my final, you're a life saver!
Thanks so much, Dr. Bazett! I can see the patterns to use to differentiate and evaluate these various series.
This guy speaks very passionitly and its amazing keep up the amazing vids homie!
you seem so proud of your progress and im over struggling and giving up
where r u now? i hope u survived it well :)
thank you!! it’s sooo hard to find concise videos like this
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!
Haven't gone to any classes and now I'm cramming for the test today.
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
I appreciate your enthusiasm.
It's It's shame I only found you the day before my Calc 2 exam. Oh well, better late than never.
I am here with 30 min left for my exam and binge watching anything that can serve as a shortcut 😭😭. Wish me luck
thank you, it really helps me prepare for my final exam !!
Best of luck!
Very clear and instructive. Thank you
watching this before my calc 2 final, thanks for helping me cook
I had trouble with determining what to use, so thank you so much for making this video!
EXACTLY WHAT I NEEDED !!
Thanks from the bottom of my heart. You truly are a legend ❤️❤️❤️
this video was a goldmine for me! Thanks a lot
Forgot all about these test. I remember I had fun with them in university :)
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.
got my ap exam today thank you🙏🏽
Thanks for saving my math... this is really helpful!
Awesome experience of studying from your lectures✨.. really very helpful💫
Seeing it from India...really nice ..
Even I want to know from where are u sir..
literally a life saver
Never have I seen a man so excited about alternating series.
Thank you for uploading this video. Helped a lot.
Wow, awesome, just wanted this explanation 👏👏
Glad to hear that!
Watching 1 day before exams thanks..
This video saved my life
Great summary and quick practice problems for the methods!
great video, super helpful and much clearer than my professor
I am here enjoying your communication skills 🥰
Hey doc very nice presentation .Keep it up..From the philippines
Fast and clear. Good job!!
awesome, cool to have a refresher for all of the tests
You have a small problem in the 4th series of the left, see that can't evaluate the sequence at n =1, because of its denominator. I still love your videos. Congrats!!
This is so helpful thank you!
great explanation, thank you!!
Truely amazing no words
Good work dr!!
Thanks my dude!
thank you sir so much you are the best of them allllll
really helpful.Thank you
Fabulous work dear
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
The quotient criteria also works for the exponential thing ❤😂
bro you made cal soooooooo easy and interesting for us..... thanks a lot.......
Thank you !! You're a saver
thank you my friend!
I love you and I love this video - thank you
So. Freaking. Helpful!
You could also include Telescoping series
Amazing content man. Very clear and helpful.
That was a great video. Love this
wow this was super helpful
at 10.53 you missed the series should be positive obligation there is 3 condition to utilize AST should be decreasing must be bigger than zero and its limit should be zero
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
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
Really good video i loved it
Well Done 😎
helped me a lot
How the heck do you get 1/25 for the second one?
Plug in 1 for n and you get 1/25
It worths the 12 minutes before exam 🙃
So helpful Thanks a lot!!!
thanks u make infinite series finite for exams (3 days down)
Great video sir. Keep it up
My exam is in 12 minutes😂 btw thank you it was helpful
Good luck lol:D
Thank u , my test went well😁
if only i could prove the convergence like this and without any rigor lol
Somebody correct me if I'm wrong. The last one you have with the denominator = n^4-n^2 diverges because "a" sub 1 DNE. You would need to start n at 2 for that to converge.