Real Analysis-convergence of a set using sandwich lemma 
I am given the problem that needed me to prove  $\frac{2^k}{k!}$ converges to $0$. 

Here is something I had in mind. I would use the sandwich lemma, which resulted in that since $\frac{1}{k!}$converges to 0, and $\frac{2^k}{3^k}$ also converges to 0, and since $$\frac{1}{k!}\leq\frac{2^k}{k!}\leq \frac{2^k}{3^k},$$ $\frac{2^k}{k!}$ converges to $0$. I am not sure if $\frac{2^k}{3^k}$ is greater than or equal to $\frac{2^k}{k!}$ however, and if it is not, I am not sure how to find a equation that is larger than $\frac{2^k}{k!}$
 A: Your argument is not correct. 
Use the fact that $k! >(3)(3)...(3)$ ($k-2$ factors) Which gives $0<\frac {2^{k}} {k!} <9(\frac 2 3)^{k}$. Apply Sandwich Theorem and note that $(\frac 2 3)^{k} \to 0$ because $\frac 2 3 <1$. 
A: You are on the right track. Following your approach, in order to apply the sandwich lemma and conclude that the required limit is zero, you should show that eventually (i.e. for all positive integer $k$ greater than some constant), the following inequality holds
$$0\leq \frac{2^k}{k!}\leq \left(\frac{2}{3}\right)^k.$$
The inequality on the left is trivial. As a lower bound it suffices to take the constant sequence $0$ instead of $1/k!$ (which is correct too).
For the inequality on the right, we prove by induction that $3^k\leq k!$ for $k\geq 7$. It is true for $k=7$, and, as regards the inductive step, for $k\geq 7$,
$$3^{k+1}=3\cdot 3^k\leq 3\cdot k!\leq (k+1)\cdot k!=(k+1)!.$$
A: A different approach to your problem. If $b_k$ is the sequence, take $$\lim\limits_{k\to\infty}\displaystyle\frac{\frac{2^{k+1}}{(k+1)!}}{\frac{2^{k}}{k!}}=\lim\limits_{k\to\infty} \frac{2^{k+1}k!}{2^{k}(k+1)!}=\lim\limits_{k\to\infty} \frac{2}{k+1}=0$$Thus, if we take $\varepsilon=1$ there exist $N\in\mathbb{N}$ such that if $n>N$ then $\frac{b_{k+1}}{b_k}<1$, i.e., $b_{k+1}<b_k$. Therefore, the sequence is eventually strictly decreasing and clearly bounded below by $0$. Thus, the sequence $b_k$ is convergent. We claim that $b_k$ converges to $0$. Suppose not, i.e., $b_k$ converges to a number $c$ with $c\neq 0$. Then the subsequence $b_{k+1}$ also converges to $c$ and then we have $$0=\lim\limits_{k\to\infty}\frac{b_{k+1}}{b_k}=\displaystyle\frac{\lim\limits_{k\to\infty} b_{k+1}}{\lim\limits_{k\to\infty} b_k}=\frac{c}{c}=1$$This is impossible. Therefore, $$\lim\limits_{k\to\infty} b_k=\lim\limits_{k\to\infty}\frac{2^k}{k!}=0$$
