2

The question is prove by induction that $2^n\le n!$ for all $n\ge 4$ So far I have completed the base case so,

Base case

$n=4$

$ 2^4\le4!$

$16\le24$

Therefore the base case holds

Inductive Step Assume F(n) is true

$2^{n+1}\le (n+1)!$

$2^{n+1}\le n!(n+1)$

$2\cdot 2^n \le n!(n+1)$ <- I don’t know what do after this to get

New Approach

$2\cdot n!\le n!(n+1)$ $ by F(n) $

$2^{n+1}\le (n+1)!$

Robo Jumble
  • 93
  • I have looked at the second answer the one with 12 votes and i dont seem to understand anything after the second line in the inductive step $2^{k+1}<2(k!)$ – Robo Jumble Mar 27 '21 at 15:01
  • is the new approach correct ? – Robo Jumble Mar 27 '21 at 15:19
  • in my initial question, i have added more steps. I am not that familiar with induction when it comes to inequalities so i dont get why there is nothing on the left handside for Daniel W Farlow's answer – Robo Jumble Mar 27 '21 at 15:25

1 Answers1

1

When $n\geqslant 4$ by assumption we have $$2^n \leqslant n! $$ Let's multiply both sides on $2$, which gives

$$ 2^{n+1} \leqslant 2 \cdot n!\quad(1)$$ now, knowing, that for $n\geqslant 4$ we have $2 \lt n+1$, we obtain

$$2 \cdot n! \lt (n+1) \cdot n! = (n+1)! \quad(2)$$

$(1)$ and $(2)$ together gives $ 2^{n+1} \lt (n+1)!$.

zkutch
  • 14,313