Greatest Lower Bound/Lower Upper Bound For Sequence

I have read almost all the questions and answers regarding the theories, definitions, and properties of the $\operatorname{glb}$ and $\operatorname{lub}$ of a sequence, but I cannot find these values for the following sequence:

$\frac{n^2}{n + 1}$

I apologize for how basic this question is, but appreciate any help.

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Notice that $\frac{n^2}{n+1}$ is approximately $n$, so you expect that the sequence has no finite least upper bound. More carefully,

$$\frac{n^2}{n+1}>\frac{n^2-1}{n+1}=\frac{(n+1)(n-1)}{n+1}=n-1\;,$$

which clearly diverges to $\infty$.

Now look at the first few terms, for $n=0,1,2$, and $3$: they are $0,\frac12,\frac43$, and $\frac94$. It appears that the sequence really is simply increasing; if so, its first term is its greatest lower bound. To check that it’s increasing, look at the difference between two consecutive terms:

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

I’ll leave it to you to finish the algebra to verify that the last fraction is always positive.

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Thank you very much for the thorough answer; it is appreciated! –  David Oct 1 '12 at 11:54
@David: You’re very welcome. –  Brian M. Scott Oct 1 '12 at 11:56

Evaluate it for some $n$'s and guess ($n=0,1,2,3..$)

What if $n\to\infty$?

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For large $n$ the fraction is large and for small $n$ it is small. The smallest $n$ in $\mathbb N$ is $0$ so the smallest value of the sequence is zero. This is the greatest lower bound since it is a lower bound and itself a value of the sequence.

Now you need similar thinking to find the least upper bound. Is the sequence bounded?

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