(we can say this because and n > 0).
now, can you think of some positive number that will work for K now?
Given a sequence a(n) = n / [(n^2)+4]
I know that:
A sequence is bounded if a K∈R exists so that the the absolute value of the sequence |a(n)| for every n∈N smaller than K is. -- |a(n)| <= K
How can i prove this? Picking a random K won't do, i guess, so how would you go about doing it?
Hey, thank you for the reply!
I would have to say K = 2, by calculating the extreme points from the derivative of the sequence..
if i had to deal just with n / n^2 i would have to say K = 1 or lower, approaching 0..
or am i approaching this problem the wrong way? :/
the point is, that n/n^2 = 1/n, and 1/n most certainly is bounded above (by 1), and also bounded below (by 0).
of course, i'm appealing to the fact that n/(n^2 + 4) is always positive, so we can drop the absolute value signs (we already have a lower bound for it, 0).
note that K = 1 is not the least upper bound for n/(n^2 + 4), but we're not being asked to find a minimal bound, any bound we find will do (as long as we can
prove it is indeed a bound). taking the derivative is over-kill, and won't work for alternating series (because (-1)^x isn't even defined for most rational x).
now, the reason why any convergent sequence is bounded, is because all but a finite number are within epsilon of the limit, and we can always find the maximum
in absolute value of a finite set, and bounds "the rest of the sequence" , so the bigger of the two is a bound for the entire sequence.
but with all due respect to Plato, we do not need to prove a sequence convergent, to prove it bounded. if we find a bound, it is bounded, whether or not it converges.
for example, the sequence: 0,1,0,1,0,1,0,1,0,1,0,1....... is bounded, but not convergent. we can choose K = 1.