# Show that the sequence $2^n/n!$ from $n=1$ to infinity is monotone decreasing.

This is actually one part of a larger problem I am solving. I have so far used Calculus to show that the sequence converges and I have made a conjecture that the limit of the sequence is 0. Now I need to show that this sequence is monotone decreasing before proving that the sequence does indeed converge to my conjectured limit of 0.

For the base case, I have that $$x_1=2$$ and $$x_2=2$$ so this follows the definition of monotone decreasing, that $$x_n \geq x_{n+1}$$. Now I am trying to use induction and I am a bit unsure how to begin this process. I tried setting $$2^n/n! \geq 2^{n+1}/(n+1)!$$ and then breaking down $$2^{n+1}/(n+1)!$$ into $$2^n/n!*2/(n+1)$$ and then multiplying both sides by $$(n+1)/2$$. I guess my goal was to show that the LHS would still be greater than the RHS after doing this.

At this point I am kind of stuck and I have no idea if this is what I should even be doing here. Any help would be greatly appreciated.

clearly for $$n \gt 1$$ $$\frac2{n+1} \lt 1 \tag{1}$$ multiplying both sides by $$\frac{2^n}{n!}$$ gives: $$\frac{2^n}{n!}\frac2{n+1} \lt \frac{2^n}{n!} 1$$ i.e. $$\frac{2^{n+1}}{(n+1)!} \lt \frac{2^n}{n!}$$
Denote $$a_n=\frac {2^n}{n!}$$, we have $$\frac{a_{n+1}}{a_n}=\frac{2^{n}n!2/(n+1)}{2^nn!}=\frac2{n+1}<1$$ when $$n>2$$.
Since $$a_n>0$$ for all $$n\in\mathbb{N}$$, $$a_{n+1}.