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Use induction to prove that $n! \leq n^{n-1}$ for all integers $n\geq 1$. I'm having a hard time with induction and my professor said this is a good future test like question if someone can post a solution and explain it would help me out a lot. Thank you.

Extremal
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user212261
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3 Answers3

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Call $n!\le n^{n-1}$ the statement $T(n)$.

$T(1)$ holds, because $1!\le 1^{1-1}=1^0=1$.

Now assume $T(n)$, i.e. $n!\le n^{n-1}$.
Then $$(n+1)!=n!(n+1)\le n^{n-1}(n+1)\le (n+1)^{n-1}(n+1)=(n+1)^n$$

So $T(n+1)$ holds too.

We've proved $T(1)$ and that $\forall n\in\mathbb N(T(n)\implies T(n+1))$. $\ \ \ \square$

user26486
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  • Another textbook question I'm stuck on says use induction to prove that for any complex number z (is not equal to) 1, and integer n(greather than or equal to) 1 , (1+z+z^2+z^3+..z^n)={1-z^(n+1)]/(z-1) – user212261 Feb 18 '15 at 21:44
  • @user212261 You mean $(1+z+z^2+\cdots+z^n)=\frac{z^{n+1}-1}{z-1}$. Observe that $\frac{z^n-1}{z-1}+z^{n}=\frac{z^{n+1}-1}{z-1}$ and clearly for $n=1$ we have that $1+z=\frac{z^2-1}{z-1}$ holds. – user26486 Feb 18 '15 at 21:52
  • yeah that seems correct the only thing is in the textbook the right hand side is 1-z^(n+1) in the numerator not the other way around but that should be equivalent – user212261 Feb 18 '15 at 21:54
  • @user212261 Let $S_n := 1+z+z^2+\cdots z^n$. Note that $zS_n = S_n + z^{n+1}-1$ and solve for $S_n$. – Bman72 Feb 18 '15 at 21:54
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    @user212261: It is better if you can post this as a question with your work by highlighting where you stuck. Just posting the question is like asking others to do your work. – Extremal Feb 18 '15 at 21:55
  • okay ill do that soon im new to the website idk how it works around here like what people do.@Mathi – user212261 Feb 18 '15 at 21:56
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    @user212261 please ask this as a separate question. Do not use comments to ask questions! – Mariano Suárez-Álvarez Feb 18 '15 at 21:57
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Induction or not, $$n! = 1\cdot 2\cdot 3\cdot\ldots\cdot n \leq 1\cdot n\cdot n\cdot\ldots\cdot n = n^{n-1}.$$

Jack D'Aurizio
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Hint $\,\ $ $\dfrac{n^{n-1}}{n!}=\, \dfrac{n}{n}\,\dfrac{n}{n\!-\!1}\cdots \dfrac{n}{3}\,\dfrac{n}{2}\, \ge\, 1,\,$ being a product of rationals $\ge 1$.

Remark $ $ This proof can be discovered purely mechanically by writing $\, {\rm f}(n)\, =\, \dfrac{n^{n-1}}{n!}\,$ as a telescoping product of its successive quotients $\rm\,f(k)/f(k\!-\!1)\,$ as below.

Multiplicative Telescopy $\ \ \rm\displaystyle f(a\!-\!1) \prod_{\large k\,=\,a}^{\large n} \dfrac{f(k)}{f(k-1)}\, =\ f(n) $

Proof $ $ Induct on $\rm\,n.\,$ Base is $\rm\, f(a\!-\!1)\frac{f(a)}{f(a-1)}=\,f(a)\,$ at $\rm\,n=a.\,$ Inductive step $\rm\,n\to n\!+\!1\,$ is

$\quad\ \displaystyle\rm f(a\!-\!1)\prod_{\large k\,=\,a}^{\large n+1}\dfrac{f(k)}{f(k\!-\!1)}\, =\, \left[f(a\!-\!1)\prod_{\large k\,=\,a}^{\large n}\dfrac{f(k)}{f(k\!-\!1)}\right]\dfrac{f(n\!+\!1)}{f(n)}\, =\, \color{brown}{f(n)}\dfrac{f(n\!+\!1)}{\color{brown}{f(n)}} \, =\, f(n\!+\!1) $

Remark $\ $ Unwinding the induction yields a vivid depiction of the telescopic cancellation

$\quad \rm\displaystyle f(a\!-\!1)\prod_{\large k\,=\,a}^{n} \frac{f(k)}{f(k\!-\!1)}\, = \ \frac{\color{#c00}{\rlap{---}f(a\!-\!1)}}{1}\frac{\color{green}{\rlap{--}f(a)}}{\color{#C00}{\rlap{---}f(a\!-\!1)}}\frac{\color{royalblue}{\rlap{---}f(a\!+\!1)}}{\color{green}{\rlap{--}f(a)}}\frac{\phantom{\rlap{--}f(3)}}{\color{royalblue}{\rlap{---}f(a\!+\!1)}}\, \cdots\, \frac{\color{brown}{\rlap{---}f(n\!-\!1)}}{\phantom{\rlap{--}f(n\!-\!1)}}\frac{f(n)}{\color{brown}{\rlap{---}f(n\!-\!1)}}\, =\ \frac{f(n)}{1} $

You can find many further examples of multiplicative telescopy in other posts here.

Community
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Bill Dubuque
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