Hello again, you are correct that the limit would be different but by the limit comparison test as long as the limit is positive and finite then both series converge or diverge. So with a limit of 1/4 or 4, both would converge or both would converge. Does that make sense? Some books only state it one way but either works. I'm happy to know you have found my video helpful. I appreciate you letting me know.
We can actually put the known convergent or divergent series in the numerator or denominator. Notice by putting 3/4^n in the numerator, it makes simplifying the ratio easier than if 3/(4^n-3) was in the numerator. However, it would probably work either way.
Because (3/4^n) is a geometric sequence with R = 1/4, which we know converses. We need to use a sequence we know converses to use the limit comparison test if we are using the limit comparison test to show convergence. In the second example, we are trying to show divergence so we use a known divergent series. I hope that helps.
@sakuranbo888 dummiesdotcom explains your question very well. "Contrary to the formal definition of the Limit Comparison Test, the limit, L, doesn’t have to be finite and positive for the test to work. If the benchmark series is convergent and the limit is zero, then your series must also converge. If the limit is infinity, you can’t conclude anything. If the benchmark series is divergent and the limit is infinity, then your series must also diverge. If the limit is zero, you learn nothing.
@ 8:09 you changed the numerator n^4+2n^2-1 to n^4+2n^2+1, same thing at 9:27 when you rewrote the final answer. Once again though, an excellent and helpful video. Thanks!
It depends on the series you are comparing to. If the limit is finite and positive, both are convergent or both are divergent. It depends if you are using the test with a known convergent or known divergent series.
@1:05, is it we can use the limit comparison test only if the direct comparison test doesn't work or we could also use it even if we know that the direct comparison test would work?
You can use it whenever the conditions are satisfied and you can easily find the limit. See the last page of this handout. www.geneseo.edu/~heap/courses/222/series_review.pdf
YES!! It does matter, i was doing problem and i got the answer to be 4 and i checked the answer and it was 1/4. So i flipped in the start i would of got the answer. Otherwise i like watching your videos, they are almost better than Khan and PatrickJMT. You are good at explaining stuff and your examples are like the ones from the Calc book. KEEP UP THE GOOD WORK!
For the outcome of the test, though, it does not matter. Both 4 and 1/4 are finite and positive, which is all this test is checking for. Indeed, if you take the limit of a_n/b_n you will always get the reciprocal of what you get if you take the limit of b_n/a_n. Whenever a number is finite and positive, its reciprocal is also finite and positive. So the test does not depend on which series goes on top.
so what happens if the limit is 0 or infinity? my textbook say nothing about it but some websites say if the limit is 0, both converges, and both diverges at infinity
for the first problem, why was (3/4^n) the A sub n rather than (3/4^n-3)? because in the second problem it was vice versa, the A sub n was (5/sqrt(n^2-1)) and B sub n was (1/n)
Because it's arbitrary what you make a sub n or b sub n. If it converges the limit will be positive and finite for either scenario regardless of order (the limits will be different though). If it diverges then the limit will be 0 or infinity. 0 is not considered finite.
yes you are absolutely correct, but you made a mistake in putting the geometric series in the numerator instead of the denominator. you need to have them switched.
is the following series convergent or divergent. My answer is convergent because limit as n approach infinty of (-1/7)^n is 0........ summation from 0 to infinty 7(-1/7)^n Just need confirmation. Thanks.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Thought I was going to fail my calc 2 test tomorrow. You are a GOD.
You are the best math instructor on UA-cam!
Hello again, you are correct that the limit would be different but by the limit comparison test as long as the limit is positive and finite then both series converge or diverge. So with a limit of 1/4 or 4, both would converge or both would converge. Does that make sense? Some books only state it one way but either works. I'm happy to know you have found my video helpful. I appreciate you letting me know.
We can actually put the known convergent or divergent series in the numerator or denominator. Notice by putting 3/4^n in the numerator, it makes simplifying the ratio easier than if 3/(4^n-3) was in the numerator. However, it would probably work either way.
Because (3/4^n) is a geometric sequence with R = 1/4, which we know converses. We need to use a sequence we know converses to use the limit comparison test if we are using the limit comparison test to show convergence. In the second example, we are trying to show divergence so we use a known divergent series. I hope that helps.
@sakuranbo888 dummiesdotcom explains your question very well. "Contrary to the formal definition of the Limit Comparison Test, the limit, L, doesn’t have to be finite and positive for the test to work. If the benchmark series is convergent and the limit is zero, then your series must also converge. If the limit is infinity, you can’t conclude anything. If the benchmark series is divergent and the limit is infinity, then your series must also diverge. If the limit is zero, you learn nothing.
@ 8:09 you changed the numerator n^4+2n^2-1 to n^4+2n^2+1, same thing at 9:27 when you rewrote the final answer. Once again though, an excellent and helpful video. Thanks!
Thank you so much, you just solved all my problems with this chapter (just in time for the test tomorrow!)
Thanks again!!!
It depends on the series you are comparing to. If the limit is finite and positive, both are convergent or both are divergent. It depends if you are using the test with a known convergent or known divergent series.
at 6:47 you said it is finite and divergent ... shouldn't it be convergent?
Thank you very much you saved my life :)
Best instructor ever!
For example #2, 5/((n^2)-1)^.5, could we use the direct comparison test as well? comparing the series to a smaller series (1/n) that diverges.
Logged in just to upvote! Great work!
@1:05, is it we can use the limit comparison test only if the direct comparison test doesn't work or we could also use it even if we know that the direct comparison test would work?
You can use it whenever the conditions are satisfied and you can easily find the limit. See the last page of this handout. www.geneseo.edu/~heap/courses/222/series_review.pdf
Actually, it doesn't matter. Either they both converge or both diverge. You can switch An and Bn
YES!! It does matter, i was doing problem and i got the answer to be 4 and i checked the answer and it was 1/4. So i flipped in the start i would of got the answer. Otherwise i like watching your videos, they are almost better than Khan and PatrickJMT. You are good at explaining stuff and your examples are like the ones from the Calc book. KEEP UP THE GOOD WORK!
For the outcome of the test, though, it does not matter. Both 4 and 1/4 are finite and positive, which is all this test is checking for. Indeed, if you take the limit of a_n/b_n you will always get the reciprocal of what you get if you take the limit of b_n/a_n. Whenever a number is finite and positive, its reciprocal is also finite and positive. So the test does not depend on which series goes on top.
@jasonjkeller79 Thank you again for noticing. Luckily it doesn't affect the limit. My writing is so small I can't annotate the change.
so what happens if the limit is 0 or infinity? my textbook say nothing about it but some websites say if the limit is 0, both converges, and both diverges at infinity
for the first problem, why was (3/4^n) the A sub n rather than (3/4^n-3)? because in the second problem it was vice versa, the A sub n was (5/sqrt(n^2-1)) and B sub n was (1/n)
Because it's arbitrary what you make a sub n or b sub n. If it converges the limit will be positive and finite for either scenario regardless of order (the limits will be different though). If it diverges then the limit will be 0 or infinity. 0 is not considered finite.
It converges due to the geometric series test |r| = 1/7 < 1.
Thank you very much sir.. The video was really helpful... thank you again
I'm glad I could help. Thank you for the comment!
yes you are absolutely correct, but you made a mistake in putting the geometric series in the numerator instead of the denominator. you need to have them switched.
is the following series convergent or divergent. My answer is convergent because limit as n approach infinty of (-1/7)^n is 0........
summation from 0 to infinty 7(-1/7)^n
Just need confirmation. Thanks.
ahh I finally understand this. Thank you so much!!!
Great! Thank you for the comment.
yes
Great explanation.
Great stuff man!
Thank you!
good explanation thank you :)
isn't that the ratio test
Thank you so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
yes he did by putting Bn/An....while it was supposed to be lim An/Bn.....
very helpful
very true
thank you
so awesome!!!!
Its is n^4+2n^2-1 not n^4+2n^2+1
@bullcleo1 Thank you so much for this...
thank uuuuuuuuuuuuuuu
he made a mistake