Prove divergence of a sequence using comparison

Question

Given $a_n > b_n$ and $\lim_\infty b_n = \infty$. Prove that $a_n \to \infty$ as well.

First of all, I use the definition of divergence for $b_n$. Thus, $\forall M\in R, \exists N s.t \forall n > N$ we have $b_n > M$ Since $a_n$ is always bigger thab $b_n$ I got $a_n > M$. Thus, $a_n$ is divergent. Is my proof correct?

Answer 1

Correct.

By divergence of $b_n$,

$$\forall M\in\mathbb R:\exists N:\forall n>N:b_n\ge M.$$

Then as $\forall n:a_n>b_n$,

$$\forall M\in\mathbb R:\exists N:\forall n>N:a_n>b_n\ge M,$$

which implies

$$\forall M\in\mathbb R:\exists N:\forall n>N:a_n\ge M.$$

Answer 2

Yes. For any $M$, there is $N$ such that $a_n>b_n>M, \forall n\geq N$