Calculus - Series4
so what's the best way to find that number? would it be putting n towards infinity or 0?
Calculus - Series4
Okay I get it now!! Thanks a lot Steve :)
Calculus - Series4
Ok so lower bound is the lowest achievable value, so upper bound will be the highest, right?
Calculus - Series4
or rather it starts off at 2
Calculus - Series4
Oh, 2 would be the upper bound? Because it can't go higher than 2?
Calculus - Series4
Because I thought that if it decreases it'll eventually get lower than 1. What is the definition of lower and upper bound, I don't get that part.
Calculus - Series4
so the upper bound is 1, since it converges, right?
Calculus - Series4
but if I put n = 1 in both I get =2 for the first and =1.8 for n+1. Or is that not how I go about doing it?
Calculus - Series4
Determine if the following sequence: { (n + 1)^2 / (n^2 + 1) }, is ascending, descending and find the lower bound b OR the upper bound B. (n + 1)^2 / (n^2 + 1) > ((n+1) + 1)^2 / ((n+1)^2 + 1) so the sequence is descending, it's the bounds I don't know how to find.
Calculus -Series3
Ok I see, I was just adding the same fractions on every line instead of just adding the answer from the previous. Thanks Steve!
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