How to prove divergent sequences?

Question

For this question, I know that the sequence diverges to infinity, but I'm not sure if I am doing it right. Here is what I have so far. Can anyone please help me out?

Determine whether the following sequence is convergent or divergent

$a_n = \{8n^3 + n^2 -2\}$

$\lim_{n \to \infty} a_n = \infty$

Wts for any $M>0$, there exists some $N>0$, st if $n>N$, then $a_n>M$.

$n^3 > N^3 > M$

$n^3(8 + \frac{1}{n} - \frac{2}{n^3}) > N^3(8 + \frac{1}{N} - \frac{2}{N^3}) > M$

$n^3(8 + \frac{1}{n} - \frac{2}{n^3}) > N > (\frac{M}{8+\frac{1}{N} -\frac{2}{N^3}})^\frac{1}{3}$

Answer 1

We fix $M>2$ and we look for $N$ such that $\forall n>N$

$$a_n = 8n^3 + n^2 -2>M$$

then choose $n=M$ and check that

$$8M^3 + M^2 -2>M\iff 8M^3 + M^2>M+2$$

which is true.

Then it suffice to choose $N\ge M$.

Answer 2

Let $M > 0$.

For any $n \ge \sqrt[3]{M+2}$ we have

$$8n^3 + n^2 - 2 \ge n^3 - 2 \ge (M+2) - 2 = M$$

Hence your sequence is unbounded.

Answer 3

A different approach: Observe that your sequence is strictly increasing and integer-valued. Thus $a_{n}\geq a_{n-1}+1$, and so $a_n\geq a_0+n$. Now you should be able to easily prove $a_n$ is unbounded and hence divergent.