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It's been 10 years since my last math class so I'm very rusty. How would I go about proving $$3^n < n!$$ where $n \geq 7$?

I understand that factorials grow faster than set values with a variable exponent. Just not sure how to start proving it mathematically.

Daniel W. Farlow
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DavidC85
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2 Answers2

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We can prove it by induction. For $n\geq 7$, let $S(n)$ denote the statement $$ S(n) : 3^n < n!. $$ Base case ($n=7$): $S(7)$ says that $3^7 = 2187<5040=7!$, and this is true.

Induction step: Fix some $k\geq 7$, and assume that $S(k)$ is true where $$ S(k) : 3^k < k! $$ To be shown is that $S(k+1)$ is true where $$ S(k+1) : 3^{k+1} < (k+1)! $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 3^{k+1} &= 3^k\cdot3\tag{by definition}\\[0.5em] &< k!\cdot 3\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &< k!\cdot (k+1)\tag{since $k\geq 7$}\\[0.5em] &= (k+1)!, \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.

By mathematical induction, the statement $S(n)$ is true for all $n\geq 7$. $\blacksquare$

Daniel W. Farlow
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  • How do you go from 3^(k+1) to 3^k times 3. I suppose I need more a more rudimentary explanation. Apologies. – DavidC85 Apr 24 '15 at 18:39
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    @DavidC85 By properties of exponents, $x^a\cdot x^b = x^{a+b}$. Going the other way, $x^{a+b}=x^a\cdot x^b$. In particular, $3^{k+1} = 3^k\cdot 3^1 = 3^k \cdot 3$ – JMoravitz Apr 24 '15 at 18:41
  • @DavidC85 Yes, what JMoravitz said, but here's another way of visualizing it: $$ \underbrace{3\cdot 3\cdot\ldots\cdot 3}{\text{$k$ times}}\cdot 3=\underbrace{3\cdot 3\cdot\ldots\cdot 3}{\text{$k+1$ times}}=3^{k+1}. $$ – Daniel W. Farlow Apr 24 '15 at 18:44
  • @DavidC85 You may like to take a look at this question from a while back that concerns exponent manipulation (trickier manipulation than what we have here). You may find some of the answers there to be useful. – Daniel W. Farlow Apr 24 '15 at 18:46
  • ok, thanks, that makes much more sense. Why does the right side go from being k! to k!*3? My professor has a very thick Chinese accent, goes very quickly and just assumes everyone gets it. – DavidC85 Apr 24 '15 at 18:47
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    @DavidC85 We have $\color{blue}{3^k}\cdot 3 < \color{blue}{ k!}\cdot 3$ because that is where the induction hypothesis is used (I've highlighted the relevant portion in blue). Are you familiar with induction as a proof technique? You may enjoy reading over this question from a while ago. Several good answers on that thread. It's a pretty tricky technique if you haven't encountered it before. – Daniel W. Farlow Apr 24 '15 at 18:50
  • We just started induction proofing, hence my confusion. I really appreciate all the help. You will probably be hearing from me more in the future. – DavidC85 Apr 24 '15 at 18:51
  • @DavidC85 Certainly! You are very welcome. I'm actually a huge fan of induction (too much so really haha). It may be beneficial for you to check out this thread of my different questions/answers with the induction tag. Lots of different problems. Seeing where other users struggled and conquered is a good way of learning the concepts yourself. Nonetheless, does my answer for this question make sense now? – Daniel W. Farlow Apr 24 '15 at 18:54
  • It is starting to. I think I understand why when its k! * (k+1) = (k+1)! That just means its the next set integer. I'm not sure on why/how the 3 turns into a (k+1) on the right side of the equation. – DavidC85 Apr 24 '15 at 19:03
  • Let us continue this discussion in chat. – DavidC85 Apr 24 '15 at 19:11
  • @DavidC85 3 do not turn into k+1. Since k >= 7, therefore bigger than 3. A positive number multplied by 3 is lesser than the same number multplied by k+1. Since we already know that 7! is bigger than 3^7, then the stament holds true. – Renato Apr 24 '15 at 19:14
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Since $2187=3^{7}<7!=5040$ then for $n>7$ $$ 3^{n}=3^{7}\cdot3^{n-7}<7!\cdot8\cdot9\cdot\ldots\cdot n=n! $$

since $8,9,\dots n>3$

Belgi
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