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