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I'm sorry if the English isn't correct, I'm not native.

Let's say for example these two:

$$\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$$

$$((-1)^n+1/n)^2$$

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The standard English terms are "bounded from above" and "bounded from below". –  Hans Lundmark Oct 31 '10 at 12:07
    
Corrected, cheers. –  Qosmo Oct 31 '10 at 14:24

1 Answer 1

up vote 2 down vote accepted

If you just want the bounded property, you can simply find crude upper and lower bounds.

E.g. if $(-1)^n+1/n$ is bounded, then so is $((-1)^n+1/n)^2$ (so we can essentially disregard the power of 2). Then we note that both $(-1)^n$ and $1/n$ are between $-1$ and $1$ for all $n \geq 1$ (I told you it was crude). Hence $-2 \leq (-1)^n+1/n \leq 2$ and $((-1)^n+1/n)^2 \leq 4$.

The first sequence $\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$ can be resolved by similar techniques.

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Thanks for the insight. –  Qosmo Oct 30 '10 at 21:38
    
I should also note, that I've assumed here that $n\geq 1$, otherwise 1/n is unbounded if n is allowed to be 0. –  Douglas S. Stones Oct 30 '10 at 21:52

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