Alternating Series Test - Calculus (KristaKingMath)

Рет қаралды 41,269

Күн бұрын

► My Sequences & Series course: www.kristaking...
Alternating Series Test calculus problem example.
● ● ● GET EXTRA HELP ● ● ●
If you could use some extra help with your math class, then check out Krista’s website // www.kristakingm...
● ● ● CONNECT WITH KRISTA ● ● ●
Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student-from basic middle school classes to advanced college calculus-figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: www.kristakingm...
FACEBOOK // / kristakingmath
TWITTER // / kristakingmath
INSTAGRAM // / kristakingmath
PINTEREST // / kristakingmath
GOOGLE+ // plus.google.co...
QUORA // www.quora.com/...

Пікірлер: 57

@kristakingmath 12 жыл бұрын

@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. :)

@rudicatiza004 11 жыл бұрын

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 12 жыл бұрын

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. :)

@kristakingmath 13 жыл бұрын

@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! :)

@geochum 7 жыл бұрын

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 7 жыл бұрын

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

@sharikgorilla176 12 жыл бұрын

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 ....

@nahidbarghi1707 2 жыл бұрын

Great video! Excellent explanation! Thanks!

@westermjean2723 5 жыл бұрын

Thank you again Krista King!

@Catofulthar 12 жыл бұрын

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?

@incorrectpajamas 6 жыл бұрын

Krista King... More like Krista QUEEN OF MATH

@gabrielcornejo4992 6 жыл бұрын

Wow! Great example.

@Nickuncle 11 жыл бұрын

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

@kristakingmath 11 жыл бұрын

Thanks for letting me know! :)

@deliamalone 12 жыл бұрын

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?

@atiquephyniatic 10 жыл бұрын

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

@kristakingmath 10 жыл бұрын

Yay! Thanks for letting me know. :)

@MrPickmybrain 12 жыл бұрын

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

@frnakiehoo123 11 жыл бұрын

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!!!

@djelite07 12 жыл бұрын

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

@craiglistly8210 11 жыл бұрын

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

@kristakingmath 11 жыл бұрын

Glad I could help! :)

@rollercoaster478 10 жыл бұрын

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???

@kristakingmath 13 жыл бұрын

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

@completeinsanity 12 жыл бұрын

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?

@ناصرالفارسي-ظ7ز 9 жыл бұрын

thank you

@kristakingmath 12 жыл бұрын

You're welcome! :)

@Youngey339 8 жыл бұрын

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

@billwindsor4224 7 жыл бұрын

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.

@completeinsanity 12 жыл бұрын

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

@joem6065 7 жыл бұрын

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 7 жыл бұрын

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.

@slobdogroof 12 жыл бұрын

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

@craiglistly8210 11 жыл бұрын

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

@OmerWins 11 жыл бұрын

Divergent I believe.

@soreloser3219 11 жыл бұрын

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

@nikburmeister758 10 жыл бұрын

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

@MrYaseen100 8 жыл бұрын

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 :/

@nigelstanford4 11 жыл бұрын

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

@kristakingmath 11 жыл бұрын

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. :)