Prove or disprove that the sequence $\{a_{n}\}$ defined by $a_{n}=3+\Big(2n-\sqrt{4n^2 +5}\Big)\sin(4n)$ is bounded. Prove or disprove that the sequence $\{a_{n}\}$ defined by $a_{n}=3+\Big(2n-\sqrt{4n^2 +5}\Big)\sin(4n)$ is bounded.
This is what I did:
First of all, we know that $|\sin(4n)|\le1 \; \forall \; n \in \Bbb N$
Then,
$$|a_{n}|=|3+\Big(2n-\sqrt{4n^2 +5}\Big)\sin(4n)|\le|3|+|2n-\sqrt{4n^2 +5}|*|\sin(4n)|\le 3+|2n-\sqrt{4n^2 +5}|$$
Also,
$$\sqrt{4n^2+5} \ge \sqrt{4n^2} \rightarrow -\sqrt{4n^2+5} \le -\sqrt{4n^2}=-2|n|=-2n \; \forall \; n \in \Bbb N$$
So, $$a_{n} \le3 +1(2n-2n)=3 \rightarrow a_{n} \le 3$$ but I don´t know if it's lower bounded.I don't know if working with the absolute value of $a_{n}$ is a good idea. Any hints ?. Thanks in advance.
 A: First,
$|\sin(4n)| \le 1$,
not 4.
Then,
if $0 < x $,
$(1+x/2)^2
=1+x+x^2/4
\gt 1+x
$
so
$\sqrt{1+x}
\lt 1+x/2
$
so that
$\begin{array}\\
0
&\lt \sqrt{4n^2+5}-2n\\
&=2n(\sqrt{1+5/(4n^2)}-1)\\
&\lt 2n((1+5/(8n^2))-1)\\
&= 2n(5/(8n^2))\\
&= 5/(4n))\\
\end{array}
$
Therefore
$|(\sqrt{4n^2+5}-2n)\sin(4n)|
\lt |\dfrac{5}{4n}|
\to 0
$.
A: My approach to this problem is primarily intuitive.  Once I form an intuition about the problem, my formal effort would be to simply justify my intuition.
If I understand correctly, the problem only requires that the existence of lower and upper bounds be established.
$a_n$ can be re-interpreted as $3 + (b_n c_n).$ 
Here, it is immediate that as $n \to \infty, b_n \to 0^-.$ 
That is $b_n$ is approaching 0 from below.
Further, $c_n$ will always be in $[-1, 1].$ 
Thus, for any $\epsilon > 0$, it is immediate that $N$ can be identified so that $\forall n > N, |b_n| < \epsilon.$
This means that after the finite # of ($N$) terms, every term beyond that must be in $[3-\epsilon, 3 + \epsilon].$
Therefore, since the $\epsilon$ consideration applies to all but a finite number of terms, the sequence must be bounded, both above and below.
