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I'm trying to prove that $\forall n\geq 3, n^{n+1}>(n+1)^n$. I came that this is true for $n>(1+\frac{1}{n})^n$. WolphramAlpha gives $n>2.293166...$ but I failed to compute it analytically.

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    $\begingroup$ $n>(1+\frac{1}{n})^n$ is FALSE...Check for $n=1$. $\endgroup$
    – Empty
    Commented Aug 17, 2015 at 13:16
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    $\begingroup$ @S.Panja-1729 It is supposed to work with $n \ge 3$. $\endgroup$
    – Crostul
    Commented Aug 17, 2015 at 13:17
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    $\begingroup$ sequence $(1+\frac{1}{n})^n$ converges to $e$ increasingly. So the terms always less than 3. inequalitiy true for $n\geq 3$. $\endgroup$
    – guest
    Commented Aug 17, 2015 at 13:19

5 Answers 5

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You should know that $(1+\frac1n)^n$ is an increasing sequence and its limit is $e<3$.

So, for $n \ge 3$ you have $$\left( 1+\frac1n \right)^n \le e < 3 \le n$$

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Crostul's solution is probably the most straightforward way to go about it, but I'm going to do some induction because I probably have nothing better to do.

Base Case: When $n = 3$, $n > \left(1 + \frac{1}{n}\right)^n$ since $3 > \frac{64}{27}$.

Inductive Step: Suppose that $n > \left(1 + \frac{1}{n}\right)^n$. We wish to show that $n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$.

Notice that with $n \ge 3$, we have $1 + \frac{1}{n+1} < 1 + \frac{1}{n}$.

Thus, if $n > \left(1 + \frac{1}{n}\right)^n$, then clearly $n > \left(1 + \frac{1}{n+1}\right)^n$.

It follows that $n\left(1 + \frac{1}{n}\right) > \left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right) \Longleftrightarrow n+1 > \left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right)$

We see that since $1 + \frac{1}{n+1} < 1 + \frac{1}{n}$, we have $\left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right) > \left(1 + \frac{1}{n+1}\right)^{n+1}$.

Therefore, $n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$.


Since we have shown a base case for $n = 3$ and have also shown that $n > \left(1 + \frac{1}{n}\right)^n \Longrightarrow n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$, we have proven that for $n \ge 3$, $n > \left(1 + \frac{1}{n}\right)^n$

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  • $\begingroup$ Great! Since you "have nothing better to do", can you give any advise on how analytically resolve $x=(1+\frac{1}{x})^x$? :) $\endgroup$
    – Lamine
    Commented Aug 17, 2015 at 13:54
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you need just to proof the inequality so you can use the binomial formula : $$ (a+1)^n=\sum_{k=0}^n C^k_n a^k $$ So : $$ \left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n C^k_n \frac{1}{n^k} $$ but $$ C^k_n\leq \frac{n^k}{k!} $$ because $$ k! C^k_n= \frac{n!}{(n-k)!}=n\times(n-1)\times\dots\times(n-k+1)\leq n^k $$ So $$ \left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n C^k_n \frac{1}{n^k}\leq \sum_{k=0}^n \frac{1}{k!}<n $$

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If you assume, inductively, that

$$\left(n\over n-1\right)^{n-1}\lt n-1$$

which holds in the base case $n-1=3$ (i.e., $(4/3)^3\lt3$), then

$$\begin{align} \left(n+1\over n\right)^n-1&=\left({n+1\over n}-1\right)\left(\left(n+1\over n\right)^{n-1}+\left(n+1\over n\right)^{n-2}+\cdots+1\right)\\ &={1\over n}\left(\left(n+1\over n\right)^{n-1}+\left(n+1\over n\right)^{n-2}+\cdots+1\right)\\ &\lt{1\over n}\left(\left(n+1\over n\right)^{n-1}+\left(n+1\over n\right)^{n-1}+\cdots+\left(n+1\over n\right)^{n-1}\right)\\ &=\left(n+1\over n\right)^{n-1}\\ &\lt\left(n\over n-1\right)^{n-1}\quad\text{(since $n^2-1\lt n^2$)}\\ &\lt n-1\quad\text{(applying the induction hypothesis here)} \end{align}$$

which implies the next inequality,

$$\left(n+1\over n\right)^n\lt n$$

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$\frac{\log(x)}{x}$ is a decreasing function for $x>e$. This implies that $\frac{\log(n+1)}{n+1}<\frac{\log(n)}{n}$ and $n\log(n+1)<(n+1)\log(n)$. The inequality follows.

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    $\begingroup$ This answer would be a lot more useful if you gave some idea of how to use this to answer the question that was asked. $\endgroup$
    – TravisJ
    Commented Aug 17, 2015 at 14:08
  • $\begingroup$ Hi, I just meant to give a hint. $\endgroup$
    – M. T
    Commented Aug 17, 2015 at 14:13

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