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.

  • $\begingroup$ If you know that the terms in your sequence are positive, then your general result holds. If $\frac{a_n}{b_n}$, while going towards zero, changes sign from time to time, then the result does not hold. $\endgroup$ – Arthur Jan 23 '16 at 17:34

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 $-\infty$.


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