Alternating Series Test - Calculus (KristaKingMath)

[@sharikgorilla176]

instead of comparing k and k+1 why dont we just compare A1 and A2? its the same thing right? so we just look at whether the previous term is larger than the next A1 > A2 > A3 ....

[@kristakingmath]

No, it's not an indeterminate form, but if you really want to prove that the limit is equal to zero, you have to multiply by the conjugate before you evaluate as k goes to infinity. If you don't do that last step, then, like you said, the square root of infinity+1 minus the square root of infinity = 0 is just an intuition. :)

[@rudicatiza004]

I just wanted to thank you! for your wonderful videos that are incredibly helpful, and i too wanted to say that the subtitles are really helpful as well for people like me, which English is not our first languaje. Thank you!

[@kristakingmath]

@Flodarelty Yes! Either search for "Limits - Conjugate Method, integralCALC" on KZbin, or you can find them on my website, here: integralcalc(dot)com/limits/, under Conjugate Method. :)

[@kristakingmath]

@Nicola72av I'm good too, and had a great weekend! :) Glad to hear that you're getting to apply some of the things you've already learned, and that you're finding the new material to be interesting! Big hugs! :)

[@completeinsanity]

No, since it's 1/sqrt(infinity)-1/sqrt(infinity +1) which goes to 0 so it doesn't require the last step. 1/(any variable) as it goes to infinity will always be 0 right?

[@incorrectpajamas]

Krista King... More like Krista QUEEN OF MATH

[@rollercoaster478]

Thank You for the explanation. Only one question: We could have divided by k^2 and then put infinity under the square root to get 0-0=0, instead of multiplying by the conjugate???

[@completeinsanity]

Is it always for all n? What if a(n+1)

[@geochum]

Hello Krista, your videos are amazing. Your intuition for high level math as well as your ability to convey difficult concepts is much appreciated. Would you consider starting a series on statistics?

[@kristakingmath]

Thanks George! Yes, I hope to start that this summer. :)

[@Catofulthar]

Loved your explanations, however using the complex conjugate to find the limit is a new idea to me. Do you happen to have a video explaining that one?

[@Youngey339]

So how do you prove that it converges absolutely/conditionally?

[@billwindsor4224]

Hey +Zqyiomgsa , to prove Absolute Convergence versus Conditional Convergence: (1) By our testing above for convergence using the alternating sign version of the function, we proved conditional convergence. (2) Next, test for convergence using the *absolute value* of the function -- if that converges, then the function converges absolutely. (3) If the alternating sign version of the function converges, but the absolute value of the function does not converge, then the function converges "only" conditionally -- that is, to achieve convergence we require the 'condition' that the function has alternating positive and negative signs.

[@deliamalone]

Awesome, thanks for taking the time to spell out the little steps, some days my brain just doesn't work and I need that!! But I do have one question: At the end it seems that you took something rather obvious, that sqrt(infinity+1) minus sqrt(infinity) would have a limit of zero and made it equal to 1/infinity would have a limit of zero. The second was a proof and the first was intution....is that because infinity-infinity is an indeterminate form?

[@joem6065]

only thing I'm wondering is why we couldn't stop after taking the derivative and plug in a value at that point to see if F prime was less than zero. Is there a reason for multiplying by the conjugate to combine the two pieces together?

[@kristakingmath]

You can't just plug in a value to see if the derivative is less than 0. You have to prove that the derivative is less than zero for every single value of k. The easiest way to do that was by combining the fractions, so that we could simplify the entire expression into one fraction, and thereby show that the numerator would always be negative and the denominator would always be positive, making the entire fraction always negative.

[@craiglistly8210]

if u cant meet both requirement does that make it divergent or does it tell us nothing in that case?

[@OmerWins]

Divergent I believe.

[@soreloser3219]

actually this test cannot prove divergence and you might need another test like the ratio test

[@nikburmeister758]

SoreLoser yes it can. when taking the limit and it is not equal to 0 the series diverges. that is the nth term test.

[@nahidbarghi1707]

Great video! Excellent explanation! Thanks!

[@MrPickmybrain]

You need to change the k's to x's in order to differentiate. The k values are technically constants.

[@nigelstanford4]

How did she get at 10:00 that the numerator is negative it is obviously a positive 1 left over to me...

[@kristakingmath]

because the numerator is k-(k+1). you have to distribute the negative sign across the binomial k+1, so you get k-k-1. then the k's cancel and you're left with -1. :)

[@westermjean2723]

Thank you again Krista King!

[@kristakingmath]

@Nicola72av Thanks Nicola! :) Hope you're doing great and had a great weekend! :D

[@atiquephyniatic]

Thank you! Your explanation makes it so much easier to understand :)

[@kristakingmath]

Yay! Thanks for letting me know. :)

[@frnakiehoo123]

Thanks so much, your nice voice makes me not want to kill myself as much while having to learn this stuff. Also, very helpful youtube video! I've been looking for one that would help me do this exact equation. Thanks!!!

[@gabrielcornejo4992]

Wow! Great example.

[@djelite07]

Thanks for the video! Can you please do more alternating series test please :) Thanks!

[@MrYaseen100]

I dont get it...if a sub's derivative is negative, it makes the formula rule true...im tryna think on a numberline scale how this works :/

[@Nickuncle]

Thanks. I'm not a fan of series but this definitely helps me out.

[@slobdogroof]

you used d/dx to take the derivative, but shouldn't it be with respect to n?

[@ناصرالفارسي-ظ7ز]

thank you

[@craiglistly8210]

thank you so much as usual you helped clear up my confusion. Your great!

[@kristakingmath]

Glad I could help! :)

[@kristakingmath]

Thanks for letting me know! :)

[@kristakingmath]

You're welcome! :)

[@dredreson420]

she made this more complicated than it needs to be

[@SebastiaoAfonso1989]

you are the best, thank you so much

[@kristakingmath]

you're welcome!

[@SebastiaoAfonso1989]

"Hi everyone" I love that lol, I always say it right when you say it with that beautiful voice (due to the all respect)

[@SebastiaoAfonso1989]

Thank you once again, have good one

[@laralord7609]

Thank you

[@kristakingmath]

You're welcome!

[@laldinable]

I like your voice, it's so please to listen

[@Catofulthar]

@TheIntegralCALC Thanks so much for the reply! Oh and your videos are great. Keep 'em comin!

[@guappoo7503]

Is it normal to like math?(Also I won't announce my age, but I'm a lot younger then you) :D

[@kristakingmath]

good! btw, i don't like series either!! :)

[@AzizSobirov]

I guess you right!!!)

[@TheKobe1503]

can we go on a math date =)

[@nigelstanford4]

Oh I'm dumb lol...thanks

[@AzizSobirov]

is this normal for 17 years old guy watching this?