Answer
{$a_{n}$} is a bounded sequence.
Work Step by Step
$a_{n} = \frac{2n-3}{3n+4}$
$f(x)= \frac{2x-3}{3x+4}$
$f'(x)= \frac{(3x+4)(2)-(2x-3)(3)}{(3x+4)^{2}}$
$=\frac{6x+8-6x+9}{(3x+4)^{2}}$
$=\frac{17}{(3x+4)^{2}} \gt 0$ for all $x$
Thus $f$ is an increasing function
So $a_{n} \lt a_{n+1}$ and the sequence is an increasing sequence.
$a_{1}=\frac{-1}{7}$, $a_{2}=\frac{1}{10}$, $a_{3}=\frac{3}{13}$, ...
Thus $a_{n} \geq -\frac{1}{7}$ for all $n \geq 1$
And so {$a_{n}$} is bounded below by $-\frac{1}{17}$
$\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty} \frac{2n-3}{3n+4}$
$=\lim\limits_{n \to \infty} \frac{2-\frac{3}{n}}{3+\frac{4}{n}}=\frac{2}{3}$
The sequence {$a_{n}$} is bounded above by $\frac{2}{3}$
Therefore {$a_{n}$} is a bounded sequence.
