Finding the limit of the product of two functions/sequences whose individual limits don't necessarily exist.

Question

Can we say that $\lim\limits_{n \to \infty} a_{n} b_{n} = \infty $ if one limit (either $\lim\limits_{n \to \infty} a_{n} $ or $\lim\limits_{n \to \infty} b_{n} $) doesn't exist ($\pm \infty$) and the other limit is non-zero? Because $\lim\limits_{n \to \infty} a_{n} b_{n} = \lim\limits_{n \to \infty} a_{n} \cdot \lim\limits_{n \to \infty} b_{n}$ is true only when both limits exist.

The context for this confusion:

Q) What is the domain of the power series $\sum_{n=0}^\infty n^nx^n$. (For what values of $x$ does it converge)

I saw a couple solutions online that used the Ratio test in the following way:

$$\text{If}\lim\limits_{n \to \infty} \left \lvert {a_{n+1} \over a_{n}} \right \rvert < 1 \text{ then } \sum a_{n} \text{ converges.}$$

So,

$$\lim\limits_{n \to \infty} \lvert {a_{n+1} \over a_{n}} \rvert = \lim\limits_{n \to \infty} \left \lvert{(n+1)^{n + 1} x^{n+1} \over n^{n}x^{n}} \right \rvert$$

$$ = x\lim\limits_{n \to \infty} \left \lvert{\left ({1 + {1 \over n}}\right )^{n} (n+1)}\right \rvert$$

Now, $\lim\limits_{n \to \infty} \left(1 +{1 \over n} \right)^n = e$ and $\lim\limits_{n \to \infty} n+1 = \infty$

The limit of the first term exists (finite non-zero number), but the limit of the second term doesn't.

So, can we conclude that $\lim\limits_{n \to \infty} \left \lvert {a_{n+1} \over a_{n}} \right \rvert = \infty$?

Answer 1

If $\lim\limits_{n\to\infty} a_n b_n$ converges and $\lim\limits_{n\to\infty} a_n$ converges (to something non-zero) then $\lim\limits_{n\to\infty} \frac{a_n b_n}{a_n} = \lim\limits_{n\to\infty} b_n$ must converge.

Hence if $\lim\limits_{n\to\infty} b_n$ does not converge and $\lim\limits_{n\to\infty} a_n$ does (and not to zero) then $\lim\limits_{n\to\infty} a_n b_n$ must also not converge.

Answer 2

Proposition: If two sequences $(a_n)_{n\in\mathbb N}$ and$(b_n)_{n\in\mathbb N}$ of real numbers are such that $\lim_{n\to\infty}a_n=+\infty$ and that $\lim_{n\to\infty}b_n=L>0$, then $\lim_{n\to\infty}(a_nb_n)=+\infty$.

Proof: Take $M>0$. There is some $p_1\in\mathbb N$ such that $n\geqslant p_1\implies a_n>\frac{2M}L$. And there is some $p_2\in\mathbb N$ such that $n\geqslant p_2\implies b_b>\frac L2$. Therefore, if $p=\max\{p_1,p_2\}$, then$$n\geqslant p\implies a_nb_n>\frac{2M}L\cdot\frac L2=M.$$