nth term divergence test | Series | AP Calculus BC | Khan Academy

[@jamal.noubani95]

I'm definitely failing Calculus II, atleast i tried, ty life gg op.

[@Newbport849]

+Jamal Al-Nubani gg

[@eminmamedov5865]

oh bro same

[@crismal6477]

Jamal Al-Nubani gg

[@DarcMagikian]

same...

[@jamal.noubani95]

@@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

[@codynichoson]

The way the letter g sits in the pink squiggle at the beginning makes me happy

[@uJonesing]

When I took this test, it said I had an aptitude for Dauntless.

[@bobbysteakhouse7022]

Nahhh

[@abcdef-zs1zq]

lmaoo

[@cankoban]

You're an amazing teacher. Thank you.

[@haarenysurendran8238]

The test didn't work on me. They call it Divergent. Thanks tester Sal Khan.

[@davidrosenman1889]

Terrific video! I was wondering about that Harmonic series. Have to watch video on it. Thanks.

[@happysingh6586]

Seen many videos to clear the logic behind this test , but now I got the reason, thank you sir ❤1

[@jasamcarcina]

Sal draws a very nice Sigma :)

[@janessa2804]

How do you know when to use this test rather than the other tests?

[@talbertvictoria5937]

so what are the conditions to apply the divergence test

[@elizabethwong7437]

Thank you so much you are fantastic

[@mrcl1043]

When we'll know that we should use more than the divergence test? Like proceed to different tests?

[@JnSubli]

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?

[@justintorres6573]

Prepare for the dauntless comments...

[@gilernt]

for the last two examples (1/n and 1/n^2), that's p-series right?

[@StricklandAssistantManager]

No, kids, not your stupid book.

[@bigbossdeleon]

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.

[@Hloo-z4w]

As per nth term it will diverge...which method is used in ur bookk

[@ben315to405thenstop]

YES I AM DIVERGENT

[@williamwang5738]

Nice work sal

[@stephaniehuynh9341]

using the test, will the series absolutely diverge when an =/ 0 ? or will there be other conditions.

[@essaitchemouthere100]

Stephanie H. For divergence, no condition. Conditions apply only for convergence.

[@evang6872]

How does 1/n not converge on 0

[@AdhirajMathur]

I'm little confused. Is this the Cauchy's fundamental test for divergence???

[@essaitchemouthere100]

Adhiraj Mathur Nope.... That's another story.

[@bronzejourney5784]

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.

[@frymckittrick6207]

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.

[@nihadnusseibeh4715]

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

[@aaronrios6470]

Nihad Nusseibeh study p-integrals 1/n does not converge as fast as 1/n^2

[@sashaburts9676]

This is so old lol but if you take another look at the improper integrals lesson they explain it

[@HardikF1]

What if the limit doesn't exist? Will the series then be divergent?

[@jga5821]

yes. For a series to converge the sequence *has to* go to 0. Nothing else works.

[@microhoarray]

what about 1 ?????? when limit goes to infinity a(n) approaches to zero YET IT DIVERGES

[@microhoarray]

OH GOD i should've wait until the end

[@althamese4494]

I don't get it. The P-series says that the last problem diverges, not converges.

[@wsar7669]

He says that it is divergence, but when the limit approaches a singular value is that not called convergence ???

[@UzumakiNaruto-xt9fm]

Calculus II is just not my thing :(

[@shannonsumpter2976]

I thought that divergence was when the limit was not finite, and convergence was when the limit was finite?

[@essaitchemouthere100]

Shannon Sumpter that's the truth but for sequence not series.This test is for Series.

[@alexandralang1088]

They call it Divergent

[@MarkVeronica-z3s]

King Glen

[@sasisakhesh4966]

When limit converges other than 0 then sigma a(sub)n doesn't diverge

[@mbenitez6722]

made some sense, not everything though :\

[@monsterhunterdude5448]

Why was this on a test for 16 year olds?

[@ClassicHolic]

Shit. I'm a Candor

[@jumbocraft6432]

Trollolololololol

[@gayboy1376]

You speak so fast