prove $a^nb^m; n<3m + 2$ is not regular by the pumping lemma

Question

I want to prove this language $a^nb^m; 0 \leq n< 3m+2$ to be not regular by the pumping lemma. This is my attempt, is this a correct way of doing it?

Let's suppose $L$ is regular. Let $s = a^{3k+1}b^{k}$ such that $k \geq 0$ as a pumping length. So we have $s \in L$ and $|s| \geq k$.

Then, by the pumping lemma, $\exists s = xyz$ such that

1. $|xy| \leq k$,

2. $|y| > 0$,

3. $\forall i \in \mathbb{N}, xy^iz \in L$

Since $|xy| \leq k$ and $|y| > 0$, then $x = a^\alpha$, $y = a^\beta (\beta \in \mathbb{N}^*)$, $z = a^{3k+1 - \alpha - \beta}b^{k}$, so $xy^iz = a^{3k +1 + i \beta - \beta}b^{k}$

Now, $xy^iz \in L \iff 3k +1 + i \beta - \beta < 3 k + 2\iff \beta(i-1) < 1$, and so $xy^iz \notin L \iff \beta(i-1) \geq 1$, which is true for $i = 2$. Thus, as this is in contradiction with the third assumption of the pumping lemma, $L$ is not regular.

Answer 1

There are two quite general classes of languages that don’t fulfil the pumping lemma: (1) a^n where n is an element of an infinite set with arbitrary large gaps between set elements. (2) a^n b^n for n >= 0.

And the generalised pumping lemma says: If L is regular then any set of substrings of elements of L fulfils the pumping lemma.

This solves practically all problems posted here very quickly. Every string a^n b^n is a substring of some string of your language, so by the generalised pumping lemma and (2) the language is not regular.