Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging $$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$
$(0\in\mathbb{N})$
Monotonicity:
To prove, that a sequence is monotonic, I can use the following inequalities:
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I inserted some $n$'s to get an idea on how the sequence is going to look like.
I got: 
\begin{align}
a_0&=3\\
a_1&=1\\
a_2&=\frac{7}{9}\approx 0.\overline{7}\\
a_3&=\frac{3}{4}=0.75
\end{align}
Assumption: The sequence is monotonic for $\forall n\in \mathbb{N}$ 
Therefore, I show that
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align} 
I am having problems when trying to prove the inequalities above:
\begin{align}
& a_n \geq a_{n+1}\Longleftrightarrow \left|\frac{a_{n+1}}{a_n}\right |\leq 1\\
& = \left|\dfrac{\dfrac{(n+1)^2+3}{(n+2)^2}}{\dfrac{n^2+3}{(n+1)^2}}\right|\\
& = \frac{4 + 10 n + 9 n^2 + 4 n^3 + n^4}{12 + 12 n + 7 n^2 + 4 n^3 + n^4}\\
& = \cdots \text{ not sure what steps I could do now}
\end{align}
Boundedness:
The upper bound with $a_n<s_o;\; s_o \in \mathbb{N}$ is obviously the first number of $\mathbb{N}$:
\begin{align}
a_0=s_o&=\frac{0^2+3}{(0+1)^2}\\
&=3
\end{align}
The lower bound $a_n>s_u;\; s_u \in \mathbb{N}$
$s_u$ should be $1$, because ${n^2+3}$ will expand similar to ${n^2+2n+1}$ when approaching infinity. I don't know how to prove that formally. 
Convergence
Assumption (s.a) $\lim_{ n \to \infty} a_n =1$
Let $\varepsilon$ contain some value, so that $\forall \varepsilon > 0\, \exists N\in\mathbb{N}\, \forall n\ge N: |a_n-a| < \varepsilon$:
\begin{align}
\mid a_n -a\mid&=\left|\frac{n^2+3}{(n+1)^2}-1\right|\\
&= \left|\frac{n^2+3}{(n+1)^2}-\left(\frac{n+1}{n+1}\right)^2\right|\\
&= \left|\frac{n^2+3-(n+1)^2}{(n+1)^2}\right|\\
&= \left|\frac{n^2+3-(n^2+2n+1)}{(n+1)^2}\right|\\
&= \left|\frac{2-2n}{(n+1)^2}\right|\\
&= \cdots \text{(how to go on?)}
\end{align}
 A: Hint: $$a_{n+1}-a_n=2\,{\frac {{n}^{2}-n-4}{ \left( n+2 \right) ^{2} \left( n+1 \right) ^{
2}}}
$$
Second hint: $$a_n=\frac{n^2(1+\frac{3}{n^2})}{n^2(1+\frac{2}{n}+\frac{1}{n^2})}$$ this tends to $1$ for $n$ tends to infinity
A: For monotonic behavior:
$a_{n+1}-a_n=\frac{(n+1)^2+3}{(n+2)^2}-\frac{n^2+3}{(n+1)^2}=\frac{(n+1)^4+3(n+1)^2-n^2(n+2)^2-3(n+2)^2}{(n^2+3n+2)^2}=\frac{2(n^2-n-4)}{(n^2+3n+2)^2}.$
Observe that for $n \geq 3$, the numerator is always positive, so for all $n \geq 3$, the sequence will be an increasing sequence (so ultimately an increasing sequence). The non-monotonic behavior only occurs in the first few terms $a_0 > a_1 > a_2>a_3<a_4<a_5<\dotsb<a_n<a_{n+1}<\dotsb$
For boundedness
\begin{align*}
a_n & = \frac{n^2+3}{(n+1)^2}\\
& \leq \frac{n^2+3}{n^2}\\
& \leq 1+\frac{3}{n^2}\\
& \leq 4 & (\forall n \geq 1)
\end{align*}
This is also satisfied by $a_0=3$.
Since it is ultimately monotonic and bounded, hence convergent.
For convergence:
You already have
\begin{align*}
|a_n-1| & = \left|\frac{2-2n}{(n+1)^2}\right|\\
& \leq \frac{2n}{(n+1)^2}\\
& \leq \frac{2n}{n^2}\\
& \leq \frac{2}{n}.
\end{align*}
For an $\epsilon >0$. Let $N$ be the smallest integer bigger than $\frac{2}{\epsilon}$. Then for all $n \geq N$, we have 
$$|a_n-1| \leq \epsilon.$$
This shows that $\lim_{n \to \infty}a_n=1$.
A: An alternative for monotonicity: 
Consider the corresponding (continuous) function $f(x) = \dfrac{x^2 + 3}{(x+1)^2}$.  You can then check (using calc 1 methods) that $f'(x) > 0$ if $x \geq 3$.  Hence $a_n$ is increasing for $n \geq 3$.  You have already checked that $a_1 \leq a_2 \leq a_3$, so the sequence is monotone increasing.
Alternatively, to show boundedness, first show that $a_n$ converges to $1$, and then use the fact that every convergent sequence is bounded (you should think about why this last fact is true).
A: In order to analyse the sequence, $$ a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$ We can look at the function, $$f(x) =\frac {x^2+3}{(x+1)^2}$$
$$f'(x) = \frac { x^2-2x-3}{(x+1)^4} >0  \text { for x>3 }$$ Therefore  the sequence is increasing for $n>3$
Note that $$\lim _{x\to \infty } f(x) =1$$  Thus $$\lim _{n\to \infty } a_n=1$$
Since every convergent sequence is bounded, so  is our sequence $$ a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$ is bounded. 
 For example we can see $|a_n|<4$ for all positive integers.
