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This question already has an answer here:

I want to prove that $(n-1)^n>n^{n-1}$, for $n>4$, $n$ is an integer.

So I divided by $n^n$ and got:

$(1-\frac{1}{n})^{n}>\frac{1}{n}$

I know that $(1-\frac{1}{n+1})^{n+1}$>$(1-\frac{1}{n})^{n+1}=(1-\frac{1}{n})^n(\frac{n-1}{n})>\frac{1}{n}(\frac{n-1}{n})$.

How can I continue?

Thanks.

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marked as duplicate by Martin Sleziak, wythagoras, Mark Bennet, Chappers, Claude Leibovici Jun 21 '15 at 11:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Solution using induction:

Base case - $n=4$, so $(4-1)^4>4^{4-1}$ which is true.

Inductive step - assume that the statement holds for some k and show that it holds for k+1, hence we want to show that $k^{k+1}>(k+1)^k$ for $k \ge 3$.

Our assumption is $(k-1)^k>k^{k-1}$ or alternatively $\displaystyle k-1>\left(1+\frac{1}{k-1}\right)^{k-1}$.

Dividing both sides by $k^k$ we get that we need to show that $\displaystyle k>\left(1+\frac{1}{k}\right)^k$ for $k \ge 3$.

It is obvious that $\displaystyle 1+\frac{1}{k}<1+\frac{1}{k-1} \rightarrow \left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{k-1}\right)^k$, so it is sufficient to show that $\displaystyle k>\left(1+\frac{1}{k-1}\right)^k$.

We can write: $$\displaystyle \left(1+\frac{1}{k-1}\right)^k=\left(1+\frac{1}{k-1}\right)\cdot{\left(1+\frac{1}{k-1}\right)^{k-1}}\underbrace{<}_{using \ assumption}\left(1+\frac{1}{k-1}\right)(k-1)=k$$

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take the logarithm of both sides and you will get $\frac{\ln(n-1)}{n-1}>\frac{\ln(n)}{n}$ to prove this consider the function $f(x)=\frac{\ln(x)}{x}$

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  • $\begingroup$ that's nice! but can i solve it by induction? $\endgroup$ – ari Nov 7 '14 at 20:17
  • $\begingroup$ Proof without induction here $\endgroup$ – M. Strochyk Nov 7 '14 at 20:30
  • $\begingroup$ @M.Strochyk - how does it help me to prove that $\frac{\ln(x-1)}{x-1}>\frac{\ln(x)}{x}$ $\endgroup$ – ari Nov 7 '14 at 20:39
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You could prove that $(1-\frac1n)^n$ is an increasing sequence from which your inequality easily follows. Just use the AM-GM inequality on $n$ copies of $\frac{n-1}n$ and the number $1$: $$1-\frac1{n+1}=\frac n{n+1}=\frac{\frac{n-1}n+\ldots+\frac{n-1}n+1}{n+1}>\sqrt[n+1]{\biggl(\frac{n-1}n\biggr)^{\!n}\!\!\cdot1}=\sqrt[n+1]{\biggl(1-\frac1n\biggr)^{\!n}}$$ Therefore $$\biggl(1-\frac1n\biggr)^{\!n}<\biggl(1-\frac1{n+1}\biggr)^{\!n+1}$$

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  • $\begingroup$ can you extend about: " from which your inequality easily follows"? $\endgroup$ – ari Nov 8 '14 at 3:32
  • $\begingroup$ @ari As you have already written in your answer, it's equivalent to $(1-\frac1n)^n>\frac1n$, so the left side is increasing and the right side decreasing, what can you tell about it? $\endgroup$ – user2345215 Nov 8 '14 at 6:41
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Another way is to take the $n(n-1)$ root of both sides. It then becomes $(n-1)^{1/(n-1)} > n^{1/n} $.

This is shown by the many proofs here that $n^{1/n}$ is a decreasing function for real $n \ge e$ or integer $n \ge 3$.

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A simple binomial expansion also does the trick. For $n\ge3$ it's true that \begin{align*}(n+1)^n&=\sum_{k=0}^n\binom nk n^{n-k}<n^n+n\cdot n^{n-1}+n(n-1)\,n^{n-2}+1+\sum_{k=3}^{n-1}n(n-1)\ldots(n-k+1)\,n^{n-k}\\&\le 3n^n+1-n^{n-1}+\sum_{k=3}^{n-1}n^kn^{n-k}<3n^n+(n-3)\,n^n=n^{n+1}.\end{align*} So your equality holds for all $n\ge 4$.

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