Absolute and conditional convergence (KristaKingMath)

[@rostyloco1]

everyone that goes online to look for conditional convergences gets stuck looking through shit loads of absolute convergence and hardly any conditional convergence. still, good vid. ty.

[@teshamartin3353]

Your videos are very helpful...Thanks for the great explanations

[@kristakingmath]

Thanks! I'm glad they're helping. :)

[@draft378]

This is exactly what I'm doing in my BC calc class thanks :)

[@brendam9374]

you are so good at explaining! such a talent!

[@kristakingmath]

+Brenda M Thank you so much!

[@hg2.]

Go Krista!

[@bibhuprasad1970]

Thank you ,for good explanation.

[@kristakingmath]

Glad you liked it!

[@muzamilanvar4564]

wow, what a useful content well explained clear voice thanks

[@kristakingmath]

You're welcome, Muzamil! I'm so glad you liked it! :)

[@olle6343]

Can you take away the absolute value bars on a function, that involves x, and see if it is only conditionally convergent on intervals of x values? Or do you recommend another test for the purpose of finding conditional convergence? Thank you!

[@HariKrishnan-bi6zt]

great explanation thank you miss

[@kristakingmath]

Thanks!

[@raajduby993]

Thank u fo well explanation of series problem

[@kristakingmath]

You're welcome, Raaj! :)

[@MrYusufgovani]

Very good explanation! Are you planning to do a course for coursera or edex ?

[@kristakingmath]

I have my own course at my website :) integralcalc.com/

[@silbridge1155]

You are beautifully amazing !!

[@mattbartlett3049]

So is L the value it converges to? If i get a value of L=0.5 is that where my series converges to?

[@kristakingmath]

No, L isn't the value it converges to. You can only tell whether or not it converges based on the value of L.

[@ramansb1008]

could you please make proof videos opposed to examples.

[@sabithasasikumar239]

I still don't get the part about 4^(2n+3), shouldn't it just be 4^(2n+2)??

[@kristakingmath]

no, because you start with 4^(2n+1), and then you plug n+1 in for n, and you get 4^[2(n+1)+1]. distributing the 2, you get 4^[2n+2+1], which simplifies to 4^(2n+3). :)

[@anthonym4366]

Those limits were not simple to do but I understand now convergent must be less than 1 .

[@muzamilanvar4564]

i study in portuguese but i understood all the steps

[@MrYaseen100]

i feel like a series will never converge conditionally, because of the absolute value, is the like the ultimate weapon lol

[@bebarshossny5766]

If i was a rich millionare i would devote every penny i have to funding you as well as khan academy trevtutor patrickjmt and every other math/science channel who literally just make videos to help students without getting anything in return Fuck my shitty professor man You people are saving my ass every damn day

[@kristakingmath]

I'm just happy that I have the opportunity to help! :D