Regular languages and the pumping lemma

Let L_n be the number of words in L of length n. If sum L_n x^n is not a rational function, then L can't be regular. See the proof in the comments.

For necessary and sufficient conditions for a language to be regular (sometimes useful in proving nonregularity when simpler tricks like the pumping lemma fail) see the Myhill–Nerode theorem.

Another good way to prove language L non-regular is to find a regular language A such that L∩A is non-regular.

For example, one can take A = a*b*, and prove that L∩A = {a^nb^n : n≥0}.

This method works because the intersection of two regular languages is always regular.