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. 
 A: 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.
A: Evaluate it for some $n$'s and guess ($n=0,1,2,3..$) 
What if $n\to\infty$?
A: 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?
