Why do I understand more from your ten minute videos than 75 minutes of class? I feel like my Calculus 2 classes show the theoretical side of what the problems will be, which is more or less interesting but doesn't seem useful to doing the homework/exams. So class ends up being basically a list of things I have to look up videos etc. about later, rather than really explaining much. Partially because of your videos I managed to get an A- on my first exam and hopefully I won't do much worse on the Series one... thanks for making these.
this is the problem: not enough hours in a day. i agree, that is how i always felt, which is why i started making all of these videos in the first place. you can not teach math at the university level without discussing the theory/proofs (that is the: WHY part) but after doing that, it leaves little time for mechanics.
You're dead on, man. Just not enough lecture time and examples for the test they give you so it's almost implied that "here's a brief overview of what we are doing, now go learn from youtube Proffs like patrickJMT."
8'th question is diverges, ı tested it with wolfram alpha too. in the end you are left with lim-->infinity 2k+2/k+3 and it is equal to 2 which is bigger than 1 and according to the ratio test it should be diverges.
The reason I like watching these videos is because you don't just work out the problems assuming we know everything, you go step by step talking through your thought process which is one of the hardest things to do in math. Thanks a lot Patrick!
eyy sir patrickjmt I just want to thank you dude, I'm 98% sure that I got a 100% on my midterm today. I watch almost all your videos about series and sequences, been watching your videos since High school and now I'm in college. Government should fund you for helping millions of students around the globe!! God bless you man!!!!
I ended up missing about 3 hours of class in this section for personal reasons and was concerned about this exam. Thanks to your awesome videos and a few long nights, I aced the exam and put myself in a great position for the final. Thank you for the help!
#8 we can solve it by using ratio test lim [@(n+1)/ @(n)]= lim [ 2(k+1) / k+1 ] when k tend to infinite equal to 2 >1 so it's Diverges .. !!!! NOT Converges
MaheR KarZouN I think you're doing too much work breaking out the ratio test; there's easier way using some reasonably simple algebra. So we want to show that [(2^k)*k!]/([(k+2)!] diverges. Recall that k! = 1 * 2 * 3 * . . . * k. By substitution and this definition, (k+1)! = 1 * 2 * 3 * . . . * k * (k+1) = k!(k+1). Similarly, (k+2)! = 1 * 2 * 3 * . . . * k * (k+1) * (k+2) = k!(k+1)(k+2). Therefore, [(2^k)*k!]/([(k+2)!] = [(2^k)*k!]/[k!(k+1)(k+2)] = [(2^k)]/[(k+1)(k+2)] = 2^k/(k^2 + 3k + 2). The limit as k approaches ininity of 2^k/(k^2 + 3k + 2) is infinity, as the numerator grows faster than the denominator. I think this is obvious, as exponentials grow faster than quadratics, but if you wanted to be more rigorous you could apply L'Hopitals rule twice. Anyway, because the limit as k approaches ininity of 2^k/(k^2 + 3k + 2) is infinity, the series diverges by the divergence test.
I had 4 hours, to learn everything I can about sequences and Series for my final. I gave up hope, until I stumbled across this amazing channel. You have no Idea how grateful I am for these videos. I love you man !
For number 4: As Patrick said, write the series like so: (-3^n* 3^1)/ 8^n. You can now combine like so: (-3/8)^n. This is now a geometric series. -3/8 is less than 1 thus it converges. First term/ 1-r. So: (9/8) / (11/8) Thus it converges to 9/11.
for number 12: divide everything by 4^k, and you get: ((5/4)^k) / ((3/4)^k +1) = infinity. since the lim k->infinity (3/4)^k =0, and lim k->infinity (5/4)^k = infinity, basically infinity/0+1 = infinity therefore the series diverges by the Test for Divergence
so great, I need to learn how to get a good grasp on series pretty quickly and just don't have the time to work the dozens of problems to get a good intuition of them like you're supposed to, this video was a lifesaver. thank you SO much.
Patrick, I would just like to thank you for putting up these videos on Sequences and Series. The praise that everybody gives you is certainly very well placed since I have learned more in six videos than I have in the last two weeks of class!
Honestly, when I come to my lecture calculus I didn't pay attention in the lecture because I am sure that every section in calculus I will find it in your fantastic videos thank you very much you are my hero
My millions of thanks will not affect you ! But i just have to thank you ! I pass calculus 1 with A grade with a (........) teacher who can't explain properly!!! And now am doing calculus 2 ... Thanks will not be enough !
why i didnt find your videos earlier those days ??? :((( iam giving tommorow morning exams and now i am trying to watch as many videos as i can you re amazing !!! thank you !!!
You're the bomb! These videos make math much clearer. Unlike some people, I love how you explain each step as to ensure our success in the topic :) And it's clearly working! I'm rocking my assignments. Thank you thank you thank you !!!
Great video. I watched it this morning before my final and it really helped me with solving problems I hadn't seen before by giving me strategies for different series/ convergence tests.
also for #12...If you break it up to (Summation) 5^k/3^k + 5^k/4^k ....you can further separate to (Summation) (5/3)^k + (5/4)^k which is the same as (Summation) (5/3)^k + (Summation) (5/4)^k ...so you have 2 geometric series with r>1, so both div. so whole diverges
On #9 you state that you can use the integral test or the direct comparison test to show convergence. You can use the integral test or the limit comparison test. but not the direct comparison test, since the series converges 1/n^2 would have to always be equal or larger, when in fact it's always a tiny bit less. Limit Comparison does the trick though. It actually gives you the same limit as the integral test.
question number 8 is divergent not convergent if you don't believe you can check that by using series calculator on internet or you can use ratio test. maybe your doctor can help you :P
Thank you! I get divergent too! I did the ratio test. I figured I should check out the comments and I'm not the only one who got divergent. The limit is going to be 2 > 1 therefore divergent
#8 is a divergent series. You can clean the a_k up to be ((2^k)/((k+1)*(k+2))), and the limit of that, as k approaches infinity, will be infinity, which is not equal to zero (0). Therefore, it is a divergent series, by the test for divergence.
@lllXchrisXlll When you pull out an n, then you should be left with 1/n^2. Basically you get rid of the + n. Then you use the 1/n^2 to compare with the original series. 1/n^2 is a convergent P series because p=2 > 1. Then use the limit comparison test.
In case you have not got an answer yet, according to wolfram alpha the series in question 8 is divergent by the limit test, however it does converge when (2^k)/(k^2+3k+2)=0.
I've been using your videos for years, I wish youtube's algorithm would but you before pages like khan academy because honestly, your content is 10x better
For #8, The series DOES diverge. When you do all the algebra (after the ratio test), you end with the lim ((2)(k+1))/(k+3) = lim 2((k+1)/(k+3)) which is roughly lim 2 (k/k) = 2...ratio test where lim >1 diverges
Just by looking at it, you can easily tell it's going to diverge. 2^k *k! is way larger than just (k+2)! Anything that is a factorial is larger than a power, a power is larger than a polynomial. So now that we have a power to the kth power and it's getting multiplied by a factorial, its growth just blows away (k+2)! Using this reasoning should be enough for you to say it'll diverge or you can take the ratio test.
I completed calc1 online with an A and I am in the process of taking Calc 2 online. I had an A in the class up until this point. Series and Sequences knocked my grade down to a C. This is extremely frustrating and makes me want to punch a baby. I can do any of these with notes and a book. As soon as I pull away the notes and book I can't complete a SINGLE ONE OF THESE. RAWRRRRRRRRRR.
no, as i said at the beginning, i was just thinking about them at a glance and could be wrong; one of them is incorrect i think, that is why i say to work them on your own! :)
Number one converges because it is a rational function. Therefore the same power over the same power, take the coefficients. Which is 1 / 1. Therefore it converges to one.
Patrick, there is a little mistake in your video. #8 does not converge. By applying the ratio test, you will get 2. Since 2>1, the series must diverge.
@Dmaqur91 i am not quite sure what you mean; the direct comparison test and limit comparison test will be inconclusive if you compare to 1/n, i do believe.
For #11 you can do direct comparison but not with 1/n, since that diverges and since its also greater than the original it doesn't say anything conclusive about the original. which is why he chose 1/n^3/2 which will be convergent by p series. since 1/n^3/2 is larger and it converges the original cannot be any greater than the convergent sum which means it also will converge.
Patrick, I've been a fan of your videos for a few years now and have to give you props for the great work. Now, I'm soo lost. Im using Stroud at the moment and i can't access the solution for this question. I'm stuck. ANY HELP AT ALL would be much appreciated. The question is prove that 1+2x/5+(3x^2)/25 +(4x^3)/125+... is convergent for -5
Hello, i think number 4 should be divergent because when you simplified it you got geometric series but you forgot to multiply (-3)^1 into (-3/8)^n. If you did, your r would be 9/8 and its greater than 1, so series diverge.
lam nguyen Hello! This is probably pointless by now, since it's a really late response, but when you have a constant multiplying inside the infinite sum, you can factor that out and you'll have simply the constant * (the whole sum). In that case: 3 * (the whole sum), so what he said is valid :)
Yes, it is tricky indeed... but with that one can't you guess it will diverge since the factorials simplify to (k+1)(k+2) in the denominator? The denominator grows as k^2 while the numerator grows as 2^k
why? do people want to see posts of what i am having for dinner? or do they want to see some selfies? savvy is the wrong word. it is just that i have zero desire to try and be some sort of celebrity. i will continue quietly making math videos instead ;)
Yeah Patrick is more like those saints who quietly drop foods in front of the houses of poor and hungry people and then leave. But, in his case people are hungry for maths.
Patrick is more like the unsung hero to the song "Big booty bitches." You know, the guy in the background; playing the keyboard and jammin' out. That dude is the key part to the song, but he's not all up in everybody's face about it.
patrickJMT I think what she was trying to say was a twitter could help those who need last minute help so they could DM you. lol I am not saying you should make one but I think thats what she was trying to say >.
2. Comparison test with 1/n^2, since 1/n^2 >= 1/n^2 + n , and 1/n^2 is convergent, the series is convergent by comparison. 3. obv alternating, prob by other ways, such as ratio test, but the alternating test is usually quicker and simpler. 4. alternating series test. (nope, he got me there. his way is much more effecient) 5. No idea what the root test is, though im sure the ratio is super simple. ) 6. perhaps comparison with 1/n^2 would help, not too sure 7. pretty sure it diverges, integral test, get integral= ln(ln n) n tends to infinity, ln x is always increasing, therefore divergent 11. comparison? 1/sqrt(n) (convergent, ratio test) > (sin1/n)/root n, therefore convergent by comparison?
Jeffrey Li on #3 if someone asks you whether it converges absolutely or conditionally, you gotta use another test other than the alternating series one, right? By the Alternating series test, it's clear that the limit (n->inf) of Bn = 0, but then you'd have to see if the absolute value of the whole series converges or not, to see if it's absolutely or conditionally converging
Álvaro Carvalho No... AST takes the abs value of Bn as n tends to infinity. If it converges by AST , that is absolute convergence. Only prob is that if it doesnt converge by AST, it doesnt mean that the series isnt conditionally convergent. Which you then have to use other tests. (Ratio tests with taking modulus, etc)
Jeffrey Li Consider An = (-1)^n * 1/n, then Bn = 1/n. If you take the absolute value of An, you'll see that the series diverges by the p-series condition (pinf) of Bn = 0, which means the alternating series An is convergent. Therefore, the AST only tells you that it either converges or it's inconclusive, but it doesn't tell you that it absolutely converges.
Great videos! I am not as fast as you but I tried to do all of them. And got pretty much most of them (well, I have to try because tomorrow is my final exam). Here are some notes on how to do each problem. Hope it helps others!! ------------------------------------------------------------------------------------------------------------------------------------------------------------------------ 1. By nth term divergent test. (1=! 0) therefore, the series diverges. (D) 2. By nth term comparison (divergent) test. (1/n^2) is a convergent series, therefore, the series may converge.(C) 3. By Alternating series test: limit of bn = 0. & bn+1
but #11 u cant use bn= 1/(n^(1/2)) as the comparison object that by p series, bn would conclude to be divergent. And bn is always bigger than or equal to sin (1/n)/(n^(1/2)).. the direction is totally wrong. comparison test would work out from other object of which that converges.
i don't think i will ever teach again most likely. without a phd, the pay typically sucks and there is no security at the university level. i only have a lowly MA degree.
TH-cam profs deserve at least half of my tuition at this point.
+StrangeAeons i can send you an address :) or you can support by giving a mere $1 per month on my patreon page :)
+patrickJMT I will definitely consider once I pass my exams!
+StrangeAeons oh, ok :)
+StrangeAeons omg hi jenny same girl
+iam kinza hahaha uw rep
time to watch this 12 times
lmfaooooooo
pro tip : you can watch series on kaldroStream. Me and my gf have been using it for watching lots of of movies these days.
@Marcelo Bodhi Definitely, have been using kaldroStream for since november myself :)
Why do I understand more from your ten minute videos than 75 minutes of class? I feel like my Calculus 2 classes show the theoretical side of what the problems will be, which is more or less interesting but doesn't seem useful to doing the homework/exams. So class ends up being basically a list of things I have to look up videos etc. about later, rather than really explaining much. Partially because of your videos I managed to get an A- on my first exam and hopefully I won't do much worse on the Series one... thanks for making these.
this is the problem: not enough hours in a day. i agree, that is how i always felt, which is why i started making all of these videos in the first place. you can not teach math at the university level without discussing the theory/proofs (that is the: WHY part) but after doing that, it leaves little time for mechanics.
You're dead on, man. Just not enough lecture time and examples for the test they give you so it's almost implied that "here's a brief overview of what we are doing, now go learn from youtube Proffs like patrickJMT."
8'th question is diverges, ı tested it with wolfram alpha too.
in the end you are left with lim-->infinity 2k+2/k+3 and it is equal to 2 which is bigger than 1 and according to the ratio test it should be diverges.
that's true!
I got that it diverges by using the ratio test
I also solved using the ratio test, and got that it diverges. I checked using wolframAlpha and it was right.
Yay got it right!
Thank god! someone else got the same as me!
series is driving me crazy, I'm slowly losing my mind
same here, feel like my brain's gonna explode
It just feels like I'm spinning my wheels.
If your brain gonna explode, r u going to "DIE-verge"?
My exam is in two days and I have just figured out it's about series :) :)
how fucked did you get
The reason I like watching these videos is because you don't just work out the problems assuming we know everything, you go step by step talking through your thought process which is one of the hardest things to do in math. Thanks a lot Patrick!
eyy sir patrickjmt I just want to thank you dude, I'm 98% sure that I got a 100% on my midterm today. I watch almost all your videos about series and sequences, been watching your videos since High school and now I'm in college. Government should fund you for helping millions of students around the globe!! God bless you man!!!!
I ended up missing about 3 hours of class in this section for personal reasons and was concerned about this exam. Thanks to your awesome videos and a few long nights, I aced the exam and put myself in a great position for the final. Thank you for the help!
#8 we can solve it by using ratio test lim [@(n+1)/ @(n)]= lim [ 2(k+1) / k+1 ] when k tend to infinite equal to 2 >1
so it's Diverges .. !!!! NOT Converges
MaheR KarZouN I think you're doing too much work breaking out the ratio test; there's easier way using some reasonably simple algebra.
So we want to show that [(2^k)*k!]/([(k+2)!] diverges. Recall that k! = 1 * 2 * 3 * . . . * k.
By substitution and this definition, (k+1)! = 1 * 2 * 3 * . . . * k * (k+1) = k!(k+1).
Similarly, (k+2)! = 1 * 2 * 3 * . . . * k * (k+1) * (k+2) = k!(k+1)(k+2).
Therefore, [(2^k)*k!]/([(k+2)!] = [(2^k)*k!]/[k!(k+1)(k+2)] = [(2^k)]/[(k+1)(k+2)] = 2^k/(k^2 + 3k + 2).
The limit as k approaches ininity of 2^k/(k^2 + 3k + 2) is infinity, as the numerator grows faster than the denominator. I think this is obvious, as exponentials grow faster than quadratics, but if you wanted to be more rigorous you could apply L'Hopitals rule twice.
Anyway, because the limit as k approaches ininity of 2^k/(k^2 + 3k + 2) is infinity, the series diverges by the divergence test.
I had 4 hours, to learn everything I can about sequences and Series for my final. I gave up hope, until I stumbled across this amazing channel. You have no Idea how grateful I am for these videos. I love you man !
8 diverges because the ratio test gives you 2>1
Thank you, from the future
the fact that you take timeout of your day to help us struggling in BC calc is fantastic! thanks a million
For number 4:
As Patrick said, write the series like so:
(-3^n* 3^1)/ 8^n.
You can now combine like so:
(-3/8)^n. This is now a geometric series. -3/8 is less than 1 thus it converges.
First term/ 1-r. So:
(9/8) / (11/8)
Thus it converges to 9/11.
I wish I could like this video a million times; the only one out there who so neatly and briefly combines all these strategies.
for number 12: divide everything by 4^k, and you get:
((5/4)^k) / ((3/4)^k +1) = infinity.
since the lim k->infinity (3/4)^k =0, and
lim k->infinity (5/4)^k = infinity,
basically infinity/0+1 = infinity
therefore the series diverges by the Test for Divergence
so great, I need to learn how to get a good grasp on series pretty quickly and just don't have the time to work the dozens of problems to get a good intuition of them like you're supposed to, this video was a lifesaver. thank you SO much.
Patrick, I would just like to thank you for putting up these videos on Sequences and Series. The praise that everybody gives you is certainly very well placed since I have learned more in six videos than I have in the last two weeks of class!
This man is helping me pass even 11 years later. idek if im gonna pass but he still helps
For number 11:
For now, consider the series sin(1/n)
it is patrickjmt :)
Honestly, when I come to my lecture calculus
I didn't pay attention in the lecture because I am sure that every section in calculus I will find it in your fantastic videos
thank you very much
you are my hero
This was very helpful. Hardest part is finding a start and as you run through 14 problems, it doesn't feel so bad doing the 15th or 16th problem.
My millions of thanks will not affect you ! But i just have to thank you ! I pass calculus 1 with A grade with a (........) teacher who can't explain properly!!! And now am doing calculus 2 ...
Thanks will not be enough !
why i didnt find your videos earlier those days ??? :(((
iam giving tommorow morning exams and now i am trying to watch as many videos as i can you re amazing !!! thank you !!!
You're the bomb! These videos make math much clearer. Unlike some people, I love how you explain each step as to ensure our success in the topic :) And it's clearly working! I'm rocking my assignments. Thank you thank you thank you !!!
You're a great man Patrick. I don't know if I've told you enough.. once again, to the rescue of my finals...
@mishimmy1 show it is absolutely convergent and bound the new series by 1/(n^2 + 1) which implies convergence by the comparison test and p-series.
Literally thank you so much. I'mma shout you out when I give my graduation speech
Great video. I watched it this morning before my final and it really helped me with solving problems I hadn't seen before by giving me strategies for different series/ convergence tests.
ohh dear lord ; love this one. lots of work was reduced just in 12 minutes. hats off to you dear.
On problem 8 I also got divergent.
my limit using the ratio test evaluated to be (2k+1)/(k+3) = 2 which is divergent.
VIVA JMT!
I can't believe im still watching this
after 10 years!!
also for #12...If you break it up to (Summation) 5^k/3^k + 5^k/4^k ....you can further separate to (Summation) (5/3)^k + (5/4)^k which is the same as (Summation) (5/3)^k + (Summation) (5/4)^k ...so you have 2 geometric series with r>1, so both div. so whole diverges
this is like the perfect video for my quiz tomorrow
PatrickJMT the best thing that happened to youtube since youtube! Thank you very much for making our lives a bit less stressful.
#2 is a good one for telescoping series using partial fractions.
You're the Andrew WK of math instruction. Thanks, sir. You are my go-to source.
On #9 you state that you can use the integral test or the direct comparison test to show convergence. You can use the integral test or the limit comparison test. but not the direct comparison test, since the series converges 1/n^2 would have to always be equal or larger, when in fact it's always a tiny bit less. Limit Comparison does the trick though. It actually gives you the same limit as the integral test.
question number 8 is divergent not convergent if you don't believe you can check that by using series calculator on internet or you can use ratio test.
maybe your doctor can help you :P
Thank you! I get divergent too! I did the ratio test. I figured I should check out the comments and I'm not the only one who got divergent. The limit is going to be 2 > 1 therefore divergent
#8 is a divergent series. You can clean the a_k up to be ((2^k)/((k+1)*(k+2))), and the limit of that, as k approaches infinity, will be infinity, which is not equal to zero (0). Therefore, it is a divergent series, by the test for divergence.
@lllXchrisXlll When you pull out an n, then you should be left with 1/n^2. Basically you get rid of the + n. Then you use the 1/n^2 to compare with the original series. 1/n^2 is a convergent P series because p=2 > 1. Then use the limit comparison test.
@marichich that is incorrect. showing that a series is smaller than a known divergent series does not prove that it the original also diverges.
In case you have not got an answer yet, according to wolfram alpha the series in question 8 is divergent by the limit test, however it does converge when (2^k)/(k^2+3k+2)=0.
I've been using your videos for years, I wish youtube's algorithm would but you before pages like khan academy because honestly, your content is 10x better
6:54 sounds like someone taking a really big hit of something lol
This guy is da real MVP.
For #8, The series DOES diverge. When you do all the algebra (after the ratio test), you end with the lim ((2)(k+1))/(k+3) = lim 2((k+1)/(k+3)) which is roughly lim 2 (k/k) = 2...ratio test where lim >1 diverges
Just by looking at it, you can easily tell it's going to diverge. 2^k *k! is way larger than just (k+2)! Anything that is a factorial is larger than a power, a power is larger than a polynomial. So now that we have a power to the kth power and it's getting multiplied by a factorial, its growth just blows away (k+2)! Using this reasoning should be enough for you to say it'll diverge or you can take the ratio test.
I completed calc1 online with an A and I am in the process of taking Calc 2 online. I had an A in the class up until this point. Series and Sequences knocked my grade down to a C. This is extremely frustrating and makes me want to punch a baby. I can do any of these with notes and a book. As soon as I pull away the notes and book I can't complete a SINGLE ONE OF THESE. RAWRRRRRRRRRR.
Thanks for the A on the mid term! I memorized the shapes of the series' and just was automatic on the test.
no, as i said at the beginning, i was just thinking about them at a glance and could be wrong; one of them is incorrect i think, that is why i say to work them on your own! :)
patrickJMT I don't understand number 2. The second portion of the series is (1/n) and that's harmonic series.These drive me crazy!!!haha
Alice Balayan it's a telescoping series, not harmonic. The denominator can be split into n(n+1) and at that point, use partial fraction expansion
Number one converges because it is a rational function. Therefore the same power over the same power, take the coefficients. Which is 1 / 1. Therefore it converges to one.
Ugh this test is about to be shenanigans :(
Sharingans
Thanks for this video! I used it as review the night before my calc 2 exam.
By the ratio test, #8 should diverge because the resulting limit is 2 which is greater than 1.
thanks for all your videos, you are the only reason why i survived calc ii. =]
You need to make your videos louder.
@jayorca glad you like 'em.
Lowkey, patrickJMT the GOAT
OMG!!! Thanks man!!! Everybody one Earth loves you and you know it! You are awesome sir. :)
Patrick you are a genius
This was extremely extremely helpful, great examples I appreciate it so so much!!!
Glad it was helpful!
Patrick, there is a little mistake in your video. #8 does not converge. By applying the ratio test, you will get 2. Since 2>1, the series must diverge.
if you use the ratio test you should get to a point where you have 2(k+1)/(k+3) and the limit = 2 which is greater than 1 so ya it would diverge
it's sad that the moment i got that 8 is divergent without using other people's solutions is one of my proudest this whole quarter T_T
@Dmaqur91 i am not quite sure what you mean; the direct comparison test and limit comparison test will be inconclusive if you compare to 1/n, i do believe.
@Wilhemet it is possible, i did not check them
For #11 you can do direct comparison but not with 1/n, since that diverges and since its also greater than the original it doesn't say anything conclusive about the original. which is why he chose 1/n^3/2 which will be convergent by p series. since 1/n^3/2 is larger and it converges the original cannot be any greater than the convergent sum which means it also will converge.
@jeffschroeder43 i think my intuition was wrong on one or two of them; you may read the other comments to see what others think
You can use the ratio test for Number 4 as well.
Problems 1-38 from Stewart’s Calculus, page 784
Should diverge if greater than 1, is what I think you meant to say.
This video is gold
Your videos are very helpful.
Is there a video where you write out the solution for a problem like number 11? I'm just curious as to how it would work out.
actually i have no questions, these videos too good :)
great video man. helped put a lot of it together
THANK YOU!
I thought I was going nuts because I thought the same thing,
Patrick, I've been a fan of your videos for a few years now and have to give you props for the great work. Now, I'm soo lost. Im using Stroud at the moment and i can't access the solution for this question. I'm stuck. ANY HELP AT ALL would be much appreciated. The question is prove that 1+2x/5+(3x^2)/25 +(4x^3)/125+... is convergent for -5
Hello, i think number 4 should be divergent because when you simplified it you got geometric series but you forgot to multiply (-3)^1 into (-3/8)^n. If you did, your r would be 9/8 and its greater than 1, so series diverge.
lam nguyen Hello! This is probably pointless by now, since it's a really late response, but when you have a constant multiplying inside the infinite sum, you can factor that out and you'll have simply the constant * (the whole sum). In that case: 3 * (the whole sum), so what he said is valid :)
no 8 is divergent
Yeah I got that too. Did you get 2 as your limit?
Wesley Tran I did. Checking with Wolfram Alpha...Correct. Diverges. SO HE'S WRONG :0
Yes, it is tricky indeed... but with that one can't you guess it will diverge since the factorials simplify to (k+1)(k+2) in the denominator? The denominator grows as k^2 while the numerator grows as 2^k
number 11 should be diverging. u can't use a direct comparison with that because 1/n^3/2 is less than (sin(1/n))/(n^1/2).
this makes cal II easier... thanks
Patrick, you should seriously become more social networking savy like a twitter so many people love your tutoring
why? do people want to see posts of what i am having for dinner? or do they want to see some selfies?
savvy is the wrong word. it is just that i have zero desire to try and be some sort of celebrity. i will continue quietly making math videos instead ;)
Yeah Patrick is more like those saints who quietly drop foods in front of the houses of poor and hungry people and then leave. But, in his case people are hungry for maths.
Patrick is more like the unsung hero to the song "Big booty bitches." You know, the guy in the background; playing the keyboard and jammin' out. That dude is the key part to the song, but he's not all up in everybody's face about it.
patrickJMT I think what she was trying to say was a twitter could help those who need last minute help so they could DM you. lol I am not saying you should make one but I think thats what she was trying to say >.
i would just LOVE to have thousands of people sending me math questions each night ;) who needs sleep, or a life, or an income?
they don't come with instructions. you just have to practice a bunch and try different things.
thanks guys maths is fun
2. Comparison test with 1/n^2, since 1/n^2 >= 1/n^2 + n , and 1/n^2 is convergent, the series is convergent by comparison.
3. obv alternating, prob by other ways, such as ratio test, but the alternating test is usually quicker and simpler.
4. alternating series test. (nope, he got me there. his way is much more effecient)
5. No idea what the root test is, though im sure the ratio is super simple. )
6. perhaps comparison with 1/n^2 would help, not too sure
7. pretty sure it diverges, integral test, get integral= ln(ln n) n tends to infinity, ln x is always increasing, therefore divergent
11. comparison? 1/sqrt(n) (convergent, ratio test) > (sin1/n)/root n, therefore convergent by comparison?
#6 ?? not sure how you were thinking of using 1/n^2 but ratio test would work
#11 1/sqrt(n) is a divergent p-series with p
Jeffrey Li on #3 if someone asks you whether it converges absolutely or conditionally, you gotta use another test other than the alternating series one, right? By the Alternating series test, it's clear that the limit (n->inf) of Bn = 0, but then you'd have to see if the absolute value of the whole series converges or not, to see if it's absolutely or conditionally converging
Álvaro Carvalho No... AST takes the abs value of Bn as n tends to infinity.
If it converges by AST , that is absolute convergence.
Only prob is that if it doesnt converge by AST, it doesnt mean that the series isnt conditionally convergent. Which you then have to use other tests. (Ratio tests with taking modulus, etc)
Jeffrey Li Consider An = (-1)^n * 1/n, then Bn = 1/n.
If you take the absolute value of An, you'll see that the series diverges by the p-series condition (pinf) of Bn = 0, which means the alternating series An is convergent.
Therefore, the AST only tells you that it either converges or it's inconclusive, but it doesn't tell you that it absolutely converges.
Brilliant explanations. Thanks!
really cool strategy
thanks for the video
Great videos! I am not as fast as you but I tried to do all of them. And got pretty much most of them (well, I have to try because tomorrow is my final exam). Here are some notes on how to do each problem. Hope it helps others!!
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1. By nth term divergent test. (1=! 0) therefore, the series diverges. (D)
2. By nth term comparison (divergent) test. (1/n^2) is a convergent series, therefore, the series may converge.(C)
3. By Alternating series test: limit of bn = 0. & bn+1
but #11 u cant use bn= 1/(n^(1/2)) as the comparison object that by p series, bn would conclude to be divergent. And bn is always bigger than or equal to sin (1/n)/(n^(1/2)).. the direction is totally wrong. comparison test would work out from other object of which that converges.
boy am i screwed for my final tomorrow
for 13 I ended up with 1/e that's still < 1 but how did you get 0?
#11 is convergent by the L.C.T. using the convergent p series 1/n^3/2 . lim n--> inf (sin(1/n)/rad(n))/ (1/n^3/2) = lim n--> inf rad(n)sin(1/n) = 0
#7 ?? is (D) with integral test
Thanks again for your help.
Would you be able to explain what you did with number 11?
number 8 DIverges
Very good methods
#8 diverges as the Limit is 2 according to my calculations.
Come teach at Georgia Tech, please.
i don't think i will ever teach again most likely. without a phd, the pay typically sucks and there is no security at the university level. i only have a lowly MA degree.
It's a shame that degrees can dictate so much, but you make a fair point.
patrickJMT the system is broken when people like you cant teach....
+patrickJMT I tought you had a PHD and you were engaged in research, I think you can bring a lot more to this subject :)
I'm at Georgia Tech too. Please come teach me.
Bless your soul sir.
I think #6 is divergence!
Thanks sir
this man is smart
This video was so helpful! Thank you :D