I'm having difficulty solving an exercise in my course.

The question is:

Prove that $n!\geq 2^n$.

We have to do this by induction. I started like this:

  1. The lowest natural number where the assumption is correct is $4$ as: $4! \geq 2^4 \iff 24 \ge 16$.
  2. The assumption is: $n! \ge 2^n$.

Now proof for $(n+1)$ which brings me to: $(n+1)! \ge 2^{(n+1)}$

I think I can rewrite it somehow like this:

$$ {(n+1)} \times {n!} \stackrel{\text{(definition of factorial)}}{\ge} 2^n \times 2 $$

$$ (n+1) \times 2^n \ge 2^n \times 2 $$

Then I think I can eliminate the $2^n$ and have something like this: $n+1 \ge 2$, or $n \ge 1$.

But I think I'm wrong here some where and was hoping somebody has some advice on this. How can I prove the above assumption?

Any help would be appreciated, kind regards.

  • $\begingroup$ Looks good to me. More might be said if you wrote why you think you're wrong. $\endgroup$ – joriki Oct 29 '11 at 16:23
  • $\begingroup$ Well problem is I'm not really experienced with proof by induction. I'm having trouble understanding it. If I end with $n >= 1$, how did I prove that the assumption is correct for every natural number after 4? $\endgroup$ – Floris Devriendt Oct 29 '11 at 16:25

In the induction step you want to show that if $k!\ge 2^k$ for some $k\ge 4$, then $(k+1)!\ge 2^{k+1}$. Since you already know that $4!\ge 2^4$, the principle of mathematical induction will then allow you to conclude that $n!\ge 2^n$ for all $n\ge 4$. You have all of the necessary pieces; you just need to put them together properly. Specifically, you can argue as follows.

Suppose that $k!\ge 2^k$, where $k\ge 4$; this is your induction hypothesis. Then $$\begin{align*} (k+1)! &= (k+1)k!\text{ (by the definition of factorial)}\\ &\ge (k+1)2^k\text{ (by the induction hypothesis)}\\ &> 2\cdot2^k\text{ (since }k\ge 4\text{)}\\ &= 2^{k+1}. \end{align*}$$ This completes the induction step: it shows that if $k\ge 4$, then $$k!\ge 2^k \implies (k+1)!\ge 2^{k+1}.$$

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    $\begingroup$ First of all thanks for your answer. How did you go from $\ge (k+1)2^k$ to $> 2 x 2^k$? $\endgroup$ – Floris Devriendt Oct 29 '11 at 17:01
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    $\begingroup$ @Floris: $k+1>2$, so $(k+1)2^k>2\cdot 2^k$. $\endgroup$ – Brian M. Scott Oct 29 '11 at 17:04
  • $\begingroup$ oh okay I get it, thanks! $\endgroup$ – Floris Devriendt Oct 29 '11 at 17:42
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    $\begingroup$ @mp19uy: Certainly: if $X>Y$, then it’s certainly true that $X\ge Y$. $\endgroup$ – Brian M. Scott Jul 30 '13 at 16:10
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    $\begingroup$ @mp19uy; Yes, because $X \geq Y \implies X > Y$ or $X = Y$. So if $X>Y$ is true, than $X \geq Y$ must be true, because the OR is satisfied. $\endgroup$ – Mythio Jul 30 '13 at 19:43

I am going to provide a different way of going about it--this is essentially Brian M. Scott's proof in reverse. As people have pointed out, if you can show that $n! > 2^n$, then you will have shown that $n! \geq 2^n$ (in this sense, you can think of $>$ as stronger than $\geq$ because $>$ implies $\geq$).

When considering $n! \geq 2^n$ and how you got stuck trying to work from left to right to prove the argument by induction, it may behoove you in some instances to actually work from right to left since $n! \geq 2^n$ is the exact same as $2^n \leq n!$. With this in mind, I will give a small proof that $2^n < n!$ for $n\geq 4$ (like I said, it is very similar to Brian's, but it provides a different way of going about it nonetheless that may be useful for you or others in the future).

For $n\geq 4$, denote the statement involving $n$ by $$ S(n) : 2^n<n!. $$ Base step ($n=4$): Since $2^4=16$ and $4!=24$, the statement $S(4)$ is true.

Inductive step: Fix some $k \geq 4$ and assume that $$ S(k) : 2^k < k! $$ is true. To be shown is that $$ S(k+1) : 2^{k+1} < (k+1)! $$ follows. Beginning with the left side of $S(k+1)$, \begin{align} 2^{k+1} &= 2(2^k)\\[0.5em] &< 2(k!)\tag{by $S(k)$}\\[0.5em] &< (k+1)(k!)\tag{since $k\geq 4$}\\[0.5em] &= (k+1)!, \end{align} the right side of $S(k+1)$. This concludes the inductive step $S(k)\to S(k+1)$.

Thus, by mathematical induction, for all $n\geq 4$, the inequality $S(n)$ is true.

  • $\begingroup$ Much more comprehensible to me. Thank you. $\endgroup$ – Ulad Kasach Sep 30 '15 at 19:32

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