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.