How can i prove if a sequence is bounded?

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?

thank you :)

Re: How can i prove if a sequence is bounded?

$\displaystyle \frac{n}{n^2 + 4} < \frac{n}{n^2} = \dots ?$

(we can say this because $\displaystyle n^2 < n^2 + 4$ and n > 0).

now, can you think of some positive number that will work for K now?

Re: How can i prove if a sequence is bounded?

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? :/

Re: How can i prove if a sequence is bounded?

I think that you are missing a most important concept here.

Theorem: If a sequence has a limit then that sequence is bounded.

If $\displaystyle (x_n)\to L$ then $\displaystyle \left( {\exists N} \right)$ such that $\displaystyle n \geqslant N\; \Rightarrow \;\left| {x_n - L} \right| < 1$.

From which it follows that $\displaystyle n \geqslant N\; \Rightarrow \;\left| {x_n } \right| < 1 + \left| L \right|$

Hence $\displaystyle \left( {\forall n} \right)\left[ {\left| {x_n } \right| \leqslant \sum\limits_{k = 1}^N {\left| {x_k } \right|} + 1 + \left| L \right|} \right]$

Re: How can i prove if a sequence is bounded?

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 $\displaystyle |L| + \epsilon$ 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.