I want to check as for the convergence the sequence $(a^n b^{n^2})$ for all the possible values that $a,b$ take.
I have thought the following:
We have that $\lim_{n \to +\infty} a^n=+\infty$ if $a>1$, $\lim_{n \to +\infty} a^n=0$ if $-1<a<1$, right?
What happens for $a<-1$ ?
For the $b$-term, we have the same as above. So we have to check all the possible combinations, right? How can we check the convergence of $(a^n b^{n^2})$ when we get an undefined form?
The product $a^nb^{n^2}$ can be rewritten as $(ab^n)^n$. Therefore, it suffices to discuss the value of $(ab^n)$.
First, we consider the sign of $ a$ and $b $.
(1) If $ab>0$, then $a^nb^{n^2}$ = $|a|^n|b|^{n^2}$.
(2) If $ab<0$, then the sign of $a^nb^{n^2}$ changes when $n$ increases by $1$.
$\qquad$Then $a^nb^{n^2}$ converges only when $|a|^n|b|^{n^2}\to 0$ .
Second, we may assume that $a>0,\ b>0$.
(1) When $b > 1$, $ab^n \to +\infty$, $(ab^n)^n\to +\infty$.
(2) When $b < 1$, $ab^n \to 0$, $(ab^n)^n\to 0$.
(3) When $b = 1$, $(ab^n)=a$, the convergenge of $(ab^n)^n$ equals the convergence of $a^n$.