From my problem set, I need to show that $(1+a)^n \ge 1+an$ for any $(a, n) \in [-1, \infty) \times \mathbb{N}$ by induction, and use this inequality to show that i) $a^n \rightarrow \infty$ if $a>1$ and $a^n \rightarrow 0$ if $|a|<1$, ii) $na^n \rightarrow 0$ when $|a|<1$, and iii) $\frac{a^n}{n!} \rightarrow 0$ for any real number $a>0$.
I have solved all of them. The induction part is trivial. For (i), $a^n \ge 1+(n-1)a \rightarrow \infty$ if $a>0$. When $|a|<1$, there exists some number $M$ such that $(\frac{1}{a})^n \ge 1+(n-1) \frac{1}{a} > \frac{1}{\epsilon}$ when $n \ge M$. Therefore we have shown that $a^n < \epsilon$.
For ii), we have $n \rightarrow \infty$ and $a^n \rightarrow 0$. I use L’Hopital’s theorem to show that $\lim_{n \rightarrow \infty} na^n = \lim_{n \rightarrow \infty} \frac{n}{a^{-n}}= \lim_{n \rightarrow \infty} (- \frac{1}{n})a^{n+1} = 0 $.
I got stuck in iii). I just stated that $ \sum_{n=0}^{\infty} \frac{a^n}{n!} =exp(a) $ is convergent and thus the sequence converges to zero. But obviously there must be some proof with Bernoulli’s inequality. Any other proofs that use Bernoulli’s inequality are welcome.
Probably my answer is not the desired one for the OP, where I use i) to prove iii), but I think this maybe the intention of the writer of the problem set.
Let $N=[a]$ be the largest integer such that $N\leq a$, then for sufficiently large $n$, we have $$\frac{a^n}{n!}=\frac{a^N}{N!}\frac{a^{n-N}}{(N+1)\cdots n}\leq \frac{a^N}{N!}\frac{a^{n-N}}{(N+1)^{n-N}}=\frac{(N+1)^N}{N!}\left(\frac{a}{N+1}\right)^n\to0,$$ using $\left|\frac{a}{N+1}\right|<1$ and the second part of i).