Đặt $u_n=\dfrac{3^n-1}{2^n}$
$u_{n+1}=\dfrac{3^{n+1}-1}{2^{n+1}}=\dfrac{3^n.3-1}{2^n.2}$
$u_{n+1}-u_n=\dfrac{3^n.3-1}{2^n.2}-\dfrac{3^n-1}{2^n}$
$=\dfrac{3^n.3-1-2(3^n-1)}{2^n.2}$
$=\dfrac{3^n+1}{2^n.2}>0$ (do $n>0$)
$\to u_{n+1}>u_n$
Vậy $(u_n)$ là dãy tăng