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- Опубліковано 30 лип 2024
- This calculus 2 video tutorial provides a basic introduction into the root test. If the limit of the nth root of the absolute value of the sequence as n goes to infinity is less than 1, the series converges. If it's greater than 1, the series diverges. If it's equal to 1, the root test is inconclusive.
Integral Test For Divergence:
• Calculus 2 - Integral ...
Remainder Estimate - Integral Test:
• Remainder Estimate For...
P-Series:
• P-series
Direct Comparison Test:
• Direct Comparison Test...
Limit Comparison Test:
• Limit Comparison Test
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Alternating Series Test:
• Alternating Series Test
Alternate Series Estimation Theorem:
• Alternate Series Estim...
Absolute & Conditional Convergence:
• Absolute Convergence, ...
The Ratio Test:
• Ratio Test
The Root Test:
• Root Test
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Series Tests - Practice Problems:
• Calculus 2 - Geometric...
Taylor & Maclaurin Polynomials:
• Taylor Polynomials & M...
Taylor's Remainder Theorem:
• Taylor's Remainder The...
Power Series - Interval Convergence:
• Power Series - Finding...
Power Series - Derivatives & Integrals:
• Power Series - Differe...
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I feel like I should be paying you tuition, bro. I'm learning more here than I do in class. Thanks for everything.
@Noor It is really not as much as you think. He has 500k views in the last month... that is not even 1k dollars
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I subscribe to his patron page for longer videos and to support him since he is awesome. Will subscribe until I finish calc 2. He is a gem.
@@sjoerdplat8833 He has a membership program on UA-cam and also a patreon page, and the subscription is pretty steep. Not going to lie, the amount of work and knowledge the guy has is worth very penny and I hope he's getting rich because he deserves that. Unfortunately, being in a lower income country, the prices are pretty much inaccessible, but his free videos are very useful eitherway. He's a God among us mere mortals
6:24 how did u manage to take that 2 out of the absolute value just like that, we have a 1 over n as a power :)
I saw someone else explain it, it ended up working out bc he could’ve taken the 1/2 out before applying the root test anyways
Am having an exam tomorrow and these videos on convergence test are really getting me into the zone.Thanks
I have calc2 and orgo exam consecutively this week and I am just binge watching all your videos on series and orgo. Thanks so much for existing. I am still freaking out and studying for both exams together but your videos are helping me a lot!
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calu2;(
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7:10 you made an error taking out 1/2 when it should've been 1/2^(1/n) but nevertheless the answer is the same and thank you for your help
I noticed the same error.Good for you.
lol after all this calc 2, sometimes I also forget something as basic as pemdas. I saw this and I was like wait isn’t that… illegal? But I think the reason why it works is because 1/2 can be factored out at the beginning of the problem before the root test is applied.
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yeah, that's correct.
You are right I noticed it too
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@🇰🇪
Professor Organic Chemistry Tutor, thank you for another incredible video/lecture on the Root Test in Calculus Two. Once again, the Root Test is not difficult, however students could have some problems with the basic algebra and using La Hospital's rules in Calculus. Professor Organic Chemistry Tutor, at the 6:26 minute mark in the video, you pulled out 1/2 instead of 1 divided by 2 raised to the one divided by n. The final solution, however, is one hundred percent correct. Please correct this small error in the video.
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I really like your video!! It help me a lots Thank you
Thanks man! Calculus 2 would have been so hard without you.
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Great video!! Thanks
Hello to all the people watching this video two hours before the test.
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Well explained, thank you!!
Good explaination. 👍
In that last example, it's good that break 3 to the (n+2) into 2 parts, but since 3 squared is a constant that is independent of n, just bring it out to the front of the summation. This'll save time (and reduce your chance of making a silly mistake since you're not doing any unnecessary calculations).
wonderful explanation
Perfect job, thank you !! 🤍
Your explanation is good
At 5:03 if you do the comparison test you get that it diverges by the p test since 1/n^2 is greater than 1
is it critical to carry the absolute value sign throughout or can we assume A(sub)n is always positive for since x>1
Would I be wrong to just say the lim as n->inf (n^1/n ) = 1 rather than setting up the y and going the extra mile?
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is it possible to take out the 2 at 6:29 i see it is inside the nth root ??
It should have been done in the first place. Before taking the root test he should have wrote another sum series such as 1/2 times sum n to power n over (2. To the power four n) then taking the root test would give us the correct answer.
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In the fourth example a factor of 1/2 was pulled out of the absolute value signs. This was an invalid operation, since the entire expression within the absolute value signs, including the 2 in the denominator, must be raised to the 1/n power.
thanks a lot bro
6:22
i think you cannot do that. taking out the 2 of the absolute value to the power of 1/n !
It's a constant. you can move it out and about as long as you don't forget about it. Think about this as moving constants out an integral.
@@gobsauce8891 the 1/2 should have been put to the power of 1/n. It bothered me when I saw it but it has no impact on the answer.
NnjaScalper exactly👌👌
İt can be done before the test you are right.
@@gobsauce8891 thing is it's inside a root, root n to be exact.
Great! thanks
Helpful
Thanks boss
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why did he put absolute value instead of putting the parenthesis it originally had in the 4th example? and how is it possible to take an integer out of the limit if it's inside a parenthesis? I'd be grateful if anyone gives me an answer
Yo, correct me if I’m wrong but for that last problem, when you had 3^2 couldn’t you just pull that out in front of the limit statement and still get the same value of 0, still making it convergent by the root test?
Ivan Martin Del Campo totally
was wondering the same thing ! thanks :)
At 6:24 how can we take the 2 out of the nth root so easily can someone explain?
absolutely cannot escape its a mistake . But Don't changes the answer
at 6:22 , are you allowed to move the 1/2 outside the limit when it is inside of a square root to the nth term?
Came here to say the same thing :)
Yea he can, I was confused at first too, but he raised 2^1 to the 1/n power which is still 2 which, since the power has been applied to it already, it can be moved out.
I don't think you can. If you plug in n as 2 as an example, it results with the square root of 2 divided by sixteen. After he takes the 1over2 out, it becomes 1 divided by sixteen.
You cannot. If you pull the 2 out, it would have to be before you apply the nth root. However, I do not believe that the previous step (moving the two out before applying the nth root) is what you should do. It is more proper to apply the nth root to the entire expression. In this case, though, it does not affect the answer...just something to note.
@@billyqiao2691 My bad, I was thinking of 1^(1/n)!
Was doing the test at 11:00 and I got the same answer but I did something tricky. I made both 1/n and n infinity. So that it was infinity with an exponent of zero which in my head turned into 1. Then I multiplied 1/2 by 1 to get the convergence answer. Is that an illegal math move?
Why would you move the constant 1/2 out of the 4th problem but not move the constant 9 out of the last problem
Thanks for the video, am in an exam now. Helped me cheat!!!
So are you going to change the name of your channel
Can we pull the 1/2 out of the absolute value without applying the 1/n exponent?
same here
sure
No
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You are doing wrong at 6:27 .You can't write (1/2)^(1/n) as 1/2 you must do it in the begining before the test. l am not saying you don't know it but it is wrong anyway. And people might be confused.( l really love your videos .) Answer is same anyway.
So the root test does not indicate the series is absolutely convergent or conditionally convergent? It only shows normal convergence and divergences, right?
Ruipeng Han both the root and ratio test show absolute convergent and hence convergence.
They are basically a generalisation of the comparison test , if you look at how to prove the ratio and root test it’ll seem more apparent to you. In the root and ratio test you are comparing them to the geometric series.
I have an Exam in 2 days so I am watching the whole list of videos
in example 4 you took out the 2 from the denominator even though it was under the nth root i think that was a mistake and you should have turned it into 2^(1/n) like you did with example 8 the last question when you took out 9 you still made it to the ^(1/n)
I like how he says inconclusive
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12:55 could you have taken the 9 outside of the limit? I did that and still got converges in the end.
did that too and got the same answer
Please I want video on gamma and beta functions
How did you take out the 1/2 of the square root of n in example 4?
Because he wanted n^n/16^n to be by themselves so he can cancel out the two n's in the end; the one in the square root and the one outside the square root so it can end up n/16 which makes figuring out the problem easier.
Visual look: [|n^n/2(16^n)|]^1/n ---> 1/2 lim n-> infinity [|(n^n/16^n)|]^1/n ---> 1/2 lim n-> infinity [|(n/16)^n|]^1/n (the n and 1/n cancel out) ----> 1/2 lim n-> infinity |n/16| and I think you can finish it from there. P.S. I know this is late but I still wanted to answer the best way I can :)
11:22 why the hell did you do all that for ? I solved it in less than a minute with root test and it was way simpler way than this !
I thank God for the tutor's life . please what is the tutor's name ?
It will be good if you explain when = 1 also
Your videos proved to be very helpful to me...bt I think there's a mistake ...you cannot take 1/2 out side the bracket directly...the result will be same if, done in correct manner.
Would it be possible to use the ratio test instead?
You can but it's easier to do the root test
Some times they ask for you to use root test or ratio test specifically
how come you distributed that 1/n exponent to the 9 in 13:24, but you didn't distribute the exponent to other constants like at 7:05, where you just simply pulled out a 1/2?
He did it wrong when he pulled out the 1/2 in the front because the constant is still under the 1/n root. He was suppose to do the same thing as he did in the 9 example.
my dude over here saving my degree.
6:32
Wait... Example 4, you cant possibly put 2 outside like that.
Because it's root n of 2.
the example no. 2 is convergent not divergent..
Bro, last question and question 4 both are same, then why are you doing 9^(1/n)
facts
For the last question, the original series is an alternating series since it has the (-1)^n, right? But in the end you have proved that the original series is convergent. I thought alternating series are never converging?
the modulus makes it positive so it no longer remains alternating
That second example isnt in indeterminate form, you dont need to use L'Hopitals rule
It is. Infinite/infinite = indeterminate
my hero
9^0 isn't 0.....
9^0 is 1 but he's multiplying that limit with the other limit: 3/n-1, which is 0. The two limits multiplied together (1x0) = 0 :)
Not all heros wear capes
Example 5 I dont think you had to do all of the stuff with ln. ∞^1/∞ is = 1 and 1/2*1 = 1/2 < 0, so its convergent. Both ways work though but on an exam I think it may be a waste of time
You know so many super hard topics... How in the heck do you remember how to do all this?
I wish people like you, and videos like yours went viral, instead of all the bullshit idiots out there.
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