The sum is equal to the sum of x^(2n+1)/(2n+1) - x^(2n)/2n evaluated at x=1. This second sum is the integral with respect to x of 1/(1-x^2) - 1 - x/(1-x^2) evaluated at x = 1. This expression simplifies to log(1+x) - x evaluated at x = 1, log(2) - 1.
@owlsmathАй бұрын
nice one
@reaperskyfall6691Ай бұрын
Method was good sir and please pray for my exams in December 20 then over and can happily go to home.😊
@owlsmathАй бұрын
thanks Good luck Skyfall! Wish you good luck on your exams! 😀👍👍👍
@MikeMagTechАй бұрын
The easy and quick way is typically more practical, but I often find the hard way to be more fun and educational...yes, the two really do go hand in hand even though our society likes to give people the impression that learning is a punishment. I know I am getting a bit political, but I think that is a terrible attitude that leads to moral decay.
@owlsmathАй бұрын
Hey Mike. thanks and good points. I've been kind of battling with this idea and its a tricky balance of showing the straightforward quick ways vs the "alternative methods". 👍