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