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.
@brendam93748 жыл бұрын
you are so good at explaining! such a talent!
@kristakingmath8 жыл бұрын
+Brenda M Thank you so much!
@hg2.8 жыл бұрын
Go Krista!
@olle63434 жыл бұрын
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!
@teshamartin33538 жыл бұрын
Your videos are very helpful...Thanks for the great explanations
@kristakingmath8 жыл бұрын
Thanks! I'm glad they're helping. :)
@draft37810 жыл бұрын
This is exactly what I'm doing in my BC calc class thanks :)
@MrYusufgovani10 жыл бұрын
Very good explanation! Are you planning to do a course for coursera or edex ?
@kristakingmath10 жыл бұрын
I have my own course at my website :) integralcalc.com/
@muzamilanvar45643 жыл бұрын
wow, what a useful content well explained clear voice thanks
@kristakingmath3 жыл бұрын
You're welcome, Muzamil! I'm so glad you liked it! :)
@bibhuprasad19708 жыл бұрын
Thank you ,for good explanation.
@kristakingmath8 жыл бұрын
Glad you liked it!
@mattbartlett30497 жыл бұрын
So is L the value it converges to? If i get a value of L=0.5 is that where my series converges to?
@kristakingmath7 жыл бұрын
No, L isn't the value it converges to. You can only tell whether or not it converges based on the value of L.
@HariKrishnan-bi6zt8 жыл бұрын
great explanation thank you miss
@kristakingmath8 жыл бұрын
Thanks!
@raajduby9936 жыл бұрын
Thank u fo well explanation of series problem
@kristakingmath6 жыл бұрын
You're welcome, Raaj! :)
@sabithasasikumar23911 жыл бұрын
I still don't get the part about 4^(2n+3), shouldn't it just be 4^(2n+2)??
@kristakingmath11 жыл бұрын
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). :)
@ramansb100810 жыл бұрын
could you please make proof videos opposed to examples.
@silbridge115510 жыл бұрын
You are beautifully amazing !!
@anthonym436610 жыл бұрын
Those limits were not simple to do but I understand now convergent must be less than 1 .
@muzamilanvar45643 жыл бұрын
i study in portuguese but i understood all the steps
@MrYaseen1008 жыл бұрын
i feel like a series will never converge conditionally, because of the absolute value, is the like the ultimate weapon lol
@bebarshossny57665 жыл бұрын
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
@kristakingmath5 жыл бұрын
I'm just happy that I have the opportunity to help! :D