Induction Problem, showing one sequence is greater than the other.

Question

Need help proving $$\frac{4}{n} \geq \frac{2^n}{n!}$$ for all natural numbers $n$ using induction. I need help finding an inequality that situates itself between one another.

Answer 1

For $n>1$ we have $$ \frac{n}{n+1}\geq\frac{2}{n+1}. $$ Assume $\frac{4}{n}\geq\frac{2^n}{n!}$ for some $n>1$, then multiplying sides we get $$ \frac{4}{n+1}\geq\frac{2^{n+1}}{(n+1)!}. $$

Answer 2

Note that trivial rearrangement makes this equivalent to showing $(n-1)! \geq 2^{n-2}\ \ \forall n \geq 2$. The $n=1$ case can be handled separately.

Note that $k!= k(k-1)! \geq 2\cdot(k-1)! \geq 2\cdot 2^{n-2} = 2^{k-1} \ \ \forall k\geq 2$.

Answer 3

Since induction is not compulsory as you said:

For $n=1,2$ it is obvious.

For $n\geq 3$ you have $$\frac{2^n}{n!}= \frac 21\cdot \underbrace{\frac{2\cdots 2}{2\cdots (n-1)}}_{\leq 1}\cdot \frac 2n\leq \frac 4n$$