Find the bound of the following sequences Find the bound  of the following sequences: 
$$a)\ \frac{n^2}{n^2+1}$$
$$b)\ \frac{n^3}{n+1}$$
My attempt: 
$a)$$$0<\frac{n^2}{n^2+1}<1, \forall n\in \Bbb N $$ 
$$\frac{n^2}{n^2+1}>0 \Leftrightarrow (n^2>0) \land (n^2+1)>0$$ 
which is true $\forall n\in \Bbb N$. 
$$\frac{n^2}{n^2+1}<1 \Leftrightarrow \frac{n^2}{n^2+1}-1<0 \Leftrightarrow \frac{-1}{n^2+1}<0 \Leftrightarrow n^2>-1$$
which is also true $\forall n\in\Bbb N$.
Is this correct?
Similarly I managed to show that in $b)\ \frac{n^3}{n+1}>0, \forall n\in\Bbb N$ but I don't know what to do about the upper bound. I assume it doesn't exist, but how do I prove it?
 A: $a)$ is correct, and so are the bounds.
For $b)$, it is true that $\forall n \in \mathbb N$, both $n^3 \geq 0$ and $n+1 \geq 0$ (infact, $n+1 \geq 1$) hold true. Therefore, their quotient is also  positive i.e. $\frac {n^3}{n+1} \geq 0$ is also true. 
However, to analyse the other direction, we can use the following trick:
$$
\frac{n^3}{n+1} = \frac{n^3+1}{n+1} - \frac 1{n+1} = n^2-n+1 - \frac{1}{n+1}
$$
Now, note that :$\frac 1{n+1} \leq 1$, so  $-\frac 1{n+1} \geq -1$. Finally, we can say that:
$$
n^2-n+1 - \frac{1}{n+1} \geq n^2-n+1 -1 \geq n^2-n \geq n(n-1)\geq n-1
$$
Therefore, as $n$ becomes larger, $n-1$ also becomes infinitely large. But then, we have prove that $n-1 \leq \frac{n^3}{n+1} $, so it follows  that $\frac{n^3}{n+1} $ is unbounded.
There are many tricks to show these kinds of limits, I have shown you one. 
A: Let $a_n = \frac{n^2}{n^2+1}$ be the first sequence from your question. 

A sequence $c_n$ is bounded if for all $n \in \mathbb{N}$, $$M_1 \leq c_n \leq M_2,$$ that is, bounded from below by $M_1$ and bounded from above by $M_2$.

In order to prove a statement $a_n \leq 1$, try to be a little more specific, like: $$a_n = \frac{n^2}{n^2+1} < \frac{n^2+1}{n^2+1} = 1,$$ and $$a_n = \frac{n^2}{n^2+1} > \frac{0}{n^2+1} = 0.$$ That way we have found $M_1 = 0$ and $M_2 = 1$. Can you establish a lower bound for $b_n$? And a (possibly) upper bound for $b_n$ and why?
A: for $n\ge 1$  you have $\frac{1}{2} \leq \frac{n}{n+1}$ indeed
 $1+\frac{1}{n}\leq 2$ that is $\frac{1}{1+\frac{1}{n}}\geq\frac{1}{2}$.
Then, $\frac{n^2}{2}\leq\frac{n^3}{n+1}$.
and clearly $\frac{n^2}{2}$ in unbouded
