Limits of sequences of inverse terms: $\lim_{n\rightarrow \infty} \frac{a}{b}=0 \rightarrow^? lim_{n\rightarrow \infty} \frac{b}{a} = \infty$

Question

I'm trying to show that $\sum_{n=1}^{\infty} \frac{n!}{2^n}$ diverges. Now I know that for any fixed number $k$

$$\lim_{n\rightarrow \infty} \frac{k^n}{n!}=0$$

Does the following statement hold: $$\lim_{n\rightarrow \infty} \frac{k^n}{n!}=0 \Longrightarrow \lim_{n\rightarrow \infty} \frac{n!}{k^n} = \infty$$?

If it does(or doesn't) how do I go about proving or disproving the general case: $$\lim_{n\rightarrow \infty} \frac{a}{b}=0 \ \ \Longrightarrow^?\ \ \lim_{n\rightarrow \infty} \frac{b}{a} = \infty$$ provided $a \neq 0$ and $b \neq 0?$

I found how to show $\lim_{n\rightarrow \infty} \frac{n!}{k^n} = \infty$ so no need to show that. Please and thank you.

Answer 1

Let's assume that $a_k$ is a positive sequence.

For convergence of $a_k$ to 0 you need to show for every $\varepsilon>0$ that there is a N such that $a_k<\varepsilon$ for all $k>N$. What does $a_k<\varepsilon$ tell you about $\frac{1}{a_k}$ (for every $k>N$)? That should help you to show convergence to $+\infty$.

But for general sequence, you can have convergence to 0 from below, you would not expect convergence to $+\infty$. For alternating signs you would'nt expect convergence to either $+\infty$ or $-\infty$.