Alternating Series Test
ฝัง
- เผยแพร่เมื่อ 29 มี.ค. 2018
- This calculus 2 video tutorial provides a basic introduction into the alternating series test and how to use it to determine the convergence and divergence of a series.
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
_________________________________
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
__________________________________
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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You've got this. Stop reading comments and go study to ace that test.
Or pass that test whatever your goal is
At this rate, my only chance is a C lmao.
Thanks boo
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@it'll be okay & @Sleeve Them All that was very motivating
You seriously are the greatest of all time. I had Calc 1 last semester and watched your videos instead of lecture because you explain things so clearly and do the algebra really well. Aced that class thanks to you! Now acing calculus 2 with your videos, once again. Wishing you the most success and happiness!! You have such a big impact on math students - esp ones like me who genuinely never thought I'd even get through calc 1 let alone 2!
Seems so easy when you look at his video but by the time I get to my university text book lol the difficulty of the problems are totally on another level.
right its completely diff lmao
@@sofir9400 felt that...
Univ text book is shit :'(
exactly the case :(
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I’m currently taking calculus 2 in the summer and it is brutal but these videos keep me going. They make everything more simple to learn.
Same
Why is it brutal?
@@leif1075 calculus can be pretty hardddd
@@leif1075 As someone who took Calc 1 over the summer, it's just a lot to take in at once. With jumping all around from integral techniques, physics problems, parametric/polar equations, and now series/sequences, I can imagine it would be hard to grasp all the concepts in a short amount of time.
@@leif1075 summer semesters are kinda short iirc
Just amazing, and I couldn't understand my professor's 4 hours of lectures
You're the best thing that ever happened to my college career
I was about to cry because I didn't understand a word until I watched this video
THANK YOU THREE YEARS LATER!!
Words can't expess how I'm grateful. Our teacher just reads off the presentations during lectures and I don't understand anything, but thanks to you I'm easily passing all of the quizes, homeworks and mid/end term exams. Got only the final exam left to pass, thanks a lot for saving my degree😭❤
I love all of your videos. They help me a lot. I just want to point out one thing in this video: at the 6 minutes mark, you said :”The test is failed, so the series diverges.” Alternating series test just can prove convergence, can not prove divergent. So the conclusion should be inconclusive. If the test failed, we should use other test to define convergence or divergent.
Zhou J This should be pinned as top comment so anyone learning the A.S.T would know that it only proves convergence.
This is correct
I know I'm late, but he said in the beginning of the video that if the first test fails, you can consider that as automatically diverging
Yea i picked up on that
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Thank you so much! This is extremely helpful and your demonstration is so clear!
Your videos make so much more sense than my professor! Thank you!!!
you look like a nerd
The methods are way more easier.
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10:21
n has to go from 2 here as the denominator is undefined at n = 1.
...thanks dude u r awesome ...i really understood math on ur hands
Between your videos and Paul's Notes for additional practice problems, I have an A+ in calculus 2. Thank you!
who else is watching this in preparation for the AP calculus exam
not me cuz ap calc is for nerdz.
man I love your videos. They're so clear and easy to follow. Thanks a lot!
Lol this guy knows my deep frustrations with this concept-- I COULD NEVER FIGURE OUT WHAT A-sub-n was in the function... turns out it was the quotient 1/n... when one sees a bunch of "n" variables everywhere it is hard to tell how to solve it. THANKS BRO... be nice to show ur face one day so we can profile u and make assumptions of u
I think you deserve way more than 7.88M Subs, Like seriously you explain so good I understand most of the stuff you explain (Not everything since i'm a 9 year old)
It would be more accurately described as it needs to fail the divergence test. As passing the test implies its divergent. Other than that excellent video
One of my classmates was confused over the meaning of passing or failing the test. Although it is kind of odd to say "failing" when it satisfies the conditions, but in this context, it would make sense to say failing the Divergence Test. It could be put in a way like the following: A checkpoint on the road to pick off a specific car. If a car doesn't satisfy the conditions, then it moves on -- passing the checkpoint. Whereas if a car meets the condition, it gets pulled off and fails to pass the checkpoint. I tried explaining it like that to her, but she doesn't seem to get it still. . .I mean both passing and failing makes sense in their own way.
@@eric693 I think of it like this: when a series passes a test then there is no need for another test because you have found the answer. Likewise, when you fail a test, you need to try another test to find convergence or divergence.
JMac he is correct . If u pass a drug test . What does that mean?
YES! It took me an hour to realize that he was implying two different things lol
That tripped me up for a bit! I had to go back and double check my notes; I thought I might have wrote the Divergence Test incorrectly! Once I realized he more meant that "this meets the first condition" I was able to get back on track.
Thank you so much. You got me a 5 on AP calc, and you care getting me through calc 2 in collage.
Organic Chemistry Tutor > Every college professor
Facts
Interesting: Harmonic series diverges by p-series test while alternating harmonic series converges by alternating series test.
This tripped me up a bit
This has always made little intuitive sense to me.
Reminds me of some wise words once spoken by Von Neumann: "Young man, in mathematics you don't understand things. You just get used to them."
bAy-Zesd
Because it’s conditionally convergent, the convergence is caused by the negative.
@@kylejohnson1248 I thought this video was a bit confusing because it doesn't mention convergence vs conditional convergence
best video to understand this topic
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Great video as always.
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this video actually saved my fucking life, i take cal2 and sequences/series are the last chapter before exams and they absolutely decimate me. i worked for eight hours on an assignment thats majority series and barely made any progress
thanks
not gonna lie, you should start your own virtual university. you just too good.
Must watch video for a variety of examples on each technique, topic or concept. I particularly like the cos(2pi) effect of creating an alternating series. I suppose sin(2pi) will have a similar effect.
Thanks Bro
why cant I just plug number like 1 in that question that has ln(n+1)/n+1
Professor Organic Chemistry Tutor, thank you for an excellent video/lecture on the Alternating Series Test in Calculus Two. The Alternating Series Test is not a difficult test to understand, however checking for convergence can be problematic. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
{(1-lnx)/x^2}... How this function is increasing in the interval [1,e) ?
Moreover, we get the max value of function when limit of [x] tends to 0... If I am wrong please make me correct.. Thank you
thanks bro
Can you really use L'Hôpitals rule at 5:40? I've always learned that functions need to be continuous for the derivative to exist, and normally I don't think you would say that a function defined on non compact set of ellements is continuous?
legend
thank u
Thanks, better than my prof x100
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your welcome
very useful video, Now I know Im alternating
thank you x
no problem
for the ln(n)/n finding if the function decreases, why do we say that one of the critical points is 0 when setting the denominator equal to 0 will cause an indeterminate form?
I think because its indeterminate , you then move on to the next one
so you don't test if lim n-> infinity (whole expression using alternator) = 0 before using this alternating series test?
The question second to last: (ln/n) is divergent by the direct comparison test.
No actually the test fails if we use direct comparison method as it is said that if summation an and summation bn are series of positive terms it should satisfy the condition that is
1) summation bn is convergent then only summation an is convergent.
2) if summation an is divergent then summation bn is divergent.... Nd in ln(n) / n.... 1/n diverges by the p series test but it is not following the giving condition which is said to be satisfied.
is it ok to differentiate numerator and denominator saperately?
If yes, what do we get on differentiating them separately ?
Differentiating the numerator and denominator separately is for L'Hopital's rule. When you get a limit in form inf/inf you can use L'Hopital rule to get a new function to work with.
12:38 why we didn't use p series for 1/n which is divergent?
we had assumed that an>0 but in the second example the denominator is negative for the first term which makes the fraction negative :/
4 minutes it took me to understand a concept that my lecturer spent 2 hours discussing.
why didn't you use absolute convergence after conditionally convergence?
helpful
Please what if its raised to the power n-1
and how to know the sum of series if its converge?
Thanks, in my university this topic goes under Calculus 1 still
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Can the 1st condition be met and the 2nd one fail? Are there situations like those?
both conditions need to be met in order for this series to be convergent. I know im late. lmfao.
Did the series actually decrease in the first example because I thought n=1 equals -1 not 1 like he put
cant believe im watching this video for a differential equations exam I took calc 2 an entire year ago i hate school so damn much
What if the limit is equal to 0, but it isn’t decreasing? Would it be convergent or dirvergent?
Ultra Then I Think it will be divergent Bcz it can’t be convergent since both conditions are not fulfilled
same, i was wondering this too!
Divergent, because for convergence, it must meet both conditions.
In the last problem, I tried doing divergence test and applied L'Hopital's Rule for cos(npi)/n... I then got DNE as my answer??? Wouldnt it diverge instead?
I think it's because you are supposed to get the limit of An only which is (1/n). the cos(pi*n) is like (-1)^n. You can only apply the divergence test for positive series, therefore you wont get the right answer if u take the limit including the cos(pi*n). that's why we use this alternating test
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I don't understand how my professor can teach the exact same thing, with the same visuals, and I don't get it. I watch your video, and I connect the dots. I can't believe I'm only paying for the degree while getting my education from TH-cam
for the last example: what if it was sin(n.pi) instead of cos(n.pi)
it would just be 0 then.
sin(pi)=0 , sin(2pi)=0 ...
well i guess im too late to answer this but still :D
If you apply the AST with when An=1/n, then An goes to zero and it passes that part of the test. However, we know that 1/n is the harmonic series which is known to be divergent. There seems to be a contradiction here. How can it both approach zero and be divergent? Does it truly pass the test or not?
By the AST the alternating harmonic series (-1)^n (1/n) converges as the limit of the sequence, 1/n as n approaches infinity is zero, fulfilling the first criteria. The second criteria states that each iteration of the sequence must be smaller than the predecessor, which is also true for 1/n for the AST. Since those two criteria are fulfilled, it must mean that the alternating harmonic series (-1)^n (1/n) converges.
However, the harmonic series in itself is divergent. It is true that the harmonic series 1/n approaches zero, and by the nth-term test, it doesn't fulfill the criteria for divergence. However, just because 1/n approaches zero does not mean it is automatically convergent, you must utilize other tests in order to see if the series is divergent or convergent. Imagine the nth-term as being the most general and broad filter that removes any series in which its sequence never approaches zero. For the case of harmonic series, if you take the integral of 1/n from 1 to b, in which b represents infinity, 1/n becomes ln(n). By the second FTC, the integral is represented by F(b) - F(1), in which F represents the antiderivative, ln(n). If we plug in infinity for n, realize that ln(∞) - ln(1) goes on to infinity, thus making the harmonic series 1/n divergent as it never converges to a definite value.