Do we have test solutions for oscillating aₙ such as aₙ = 1+sin(n); aₙ = sin(n), aₙ = (-1)ⁿ ? By intuition aₙ = 1+sin(n) should diverge while aₙ = (-1)ⁿ repeats infinitely between 1 and 0 so what about sin(n) where n is in radians? What test can we use to show if really will diverge or the sums just repeating differently and infinitely without diverging?
There is an example in my book where it shows bsubn = n/(1-2n) converges to -1/2. But the nth term test suggests that this series will converge? All these convergence and divergence tests are killing me.
I didn't understand why the first series is divergent none the less it approaches 1/3 as x goes to infinity Moreover why is 1/n is divergent but 1/n squared is convergent
Limit goes to 1/3, shouldnt it converge? Makes no sense. Edit: Yeah my stupidity, a sub n converges to 1/3 normally, but when we add infinite numbers of this converged a sub n's together, we got a divergent sum of things which goes to infinite. Note to self, next time dont try to outsmart the math itself.
You start by saying this is a test to see if a series converges or diverges than later say it cannot confirm convergence. Why do the authorities not see that all these simple ideas are trivial if they are not all brought into students studies all at the same second with similar names, tests, consequences, structuring, reasoning...I swear either this calc course is made to weed out the weak or... your mistake reminded me of my ...life right now.
I'm definitely failing Calculus II, atleast i tried, ty life gg op.
+Jamal Al-Nubani gg
oh bro same
Jamal Al-Nubani gg
same...
@@DarcMagikian ok so, I failed, did it one more time and passed with a D. I graduated uni last year. If I did it literally anyone can do it
The way the letter g sits in the pink squiggle at the beginning makes me happy
When I took this test, it said I had an aptitude for Dauntless.
Nahhh
lmaoo
You're an amazing teacher. Thank you.
The test didn't work on me. They call it Divergent. Thanks tester Sal Khan.
Sal draws a very nice Sigma :)
so what are the conditions to apply the divergence test
How do you know when to use this test rather than the other tests?
Terrific video! I was wondering about that Harmonic series. Have to watch video on it. Thanks.
Seen many videos to clear the logic behind this test , but now I got the reason, thank you sir ❤1
Thank you so much you are fantastic
When we'll know that we should use more than the divergence test? Like proceed to different tests?
Prepare for the dauntless comments...
Do we have test solutions for oscillating aₙ such as aₙ = 1+sin(n); aₙ = sin(n), aₙ = (-1)ⁿ ? By intuition aₙ = 1+sin(n) should diverge while aₙ = (-1)ⁿ repeats infinitely between 1 and 0 so what about sin(n) where n is in radians? What test can we use to show if really will diverge or the sums just repeating differently and infinitely without diverging?
YES I AM DIVERGENT
Nice work sal
There is an example in my book where it shows bsubn = n/(1-2n) converges to -1/2. But the nth term test suggests that this series will converge? All these convergence and divergence tests are killing me.
As per nth term it will diverge...which method is used in ur bookk
for the last two examples (1/n and 1/n^2), that's p-series right?
How does 1/n not converge on 0
using the test, will the series absolutely diverge when an =/ 0 ? or will there be other conditions.
Stephanie H. For divergence, no condition. Conditions apply only for convergence.
I'm little confused. Is this the Cauchy's fundamental test for divergence???
Adhiraj Mathur Nope.... That's another story.
What if the limit doesn't exist?
Will the series then be divergent?
yes. For a series to converge the sequence *has to* go to 0. Nothing else works.
Calculus II is just not my thing :(
I didn't understand why the first series is divergent none the less it approaches 1/3 as x goes to infinity
Moreover why is 1/n is divergent but 1/n squared is convergent
Nihad Nusseibeh study p-integrals 1/n does not converge as fast as 1/n^2
This is so old lol but if you take another look at the improper integrals lesson they explain it
Limit goes to 1/3, shouldnt it converge? Makes no sense.
Edit: Yeah my stupidity, a sub n converges to 1/3 normally, but when we add infinite numbers of this converged a sub n's together, we got a divergent sum of things which goes to infinite. Note to self, next time dont try to outsmart the math itself.
what about 1
?????? when limit goes to infinity a(n) approaches to zero YET IT DIVERGES
OH GOD i should've wait until the end
They call it Divergent
I don't get it. The P-series says that the last problem diverges, not converges.
He says that it is divergence, but when the limit approaches a singular value is that not called convergence ???
You start by saying this is a test to see if a series converges or diverges than later say it cannot confirm convergence. Why do the authorities not see that all these simple ideas are trivial if they are not all brought into students studies all at the same second with similar names, tests, consequences, structuring, reasoning...I swear either this calc course is made to weed out the weak or... your mistake reminded me of my ...life right now.
King Glen
No, kids, not your stupid book.
I thought that divergence was when the limit was not finite, and convergence was when the limit was finite?
Shannon Sumpter that's the truth but for sequence not series.This test is for Series.
made some sense, not everything though :\
When limit converges other than 0 then sigma a(sub)n doesn't diverge
Why was this on a test for 16 year olds?
Shit. I'm a Candor
Trollolololololol
You speak so fast