Answer:
Option (1)
Step-by-step explanation:
Sequence has been given as,
[tex]a_n=\frac{n^3+3n}{n^2-6n}[/tex]
[tex]\lim_{n \to \infty}\frac{n^3+3n}{n^2-6n}[/tex]
[tex]=\lim_{n \to \infty}\frac{\frac{n^3+3n}{n^3} }{\frac{n^2-6n}{n^3} }[/tex]
[tex]=\lim_{n \to \infty}\frac{1+\frac{3}{n^2}}{\frac{1}{n}-\frac{6}{n^2} }[/tex]
= ∞
That means with the increase in the value of 'n' sequence gets larger and larger.
Therefore, sequence will diverge.
Option 1 will be the correct option.
