# Prove that $n^{(n+1)} > (n+1)^{n}$ when $n\geq 3$ [closed]

Firstly, for $n=3$,

$$3^4 > 4^3$$.

Secondly, $(n+1)^{(n+2)} > (n+2)^{(n+1)}$
Now I'm stuck.

## closed as off-topic by Daniel W. Farlow, Lord_Farin, user1551, graydad, colormegoneSep 7 '15 at 17:58

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• Do you want an induction argument ? or something else suffices ? – user252450 Sep 6 '15 at 15:31
• I need to simplify that thing to prove that 2n>n, 3>4 or whatever. – Mex Sep 6 '15 at 15:33
• So you want to prove it by induction ? – user252450 Sep 6 '15 at 15:36
• Yes, please. Mathematical induction. – Mex Sep 6 '15 at 15:38

Let's generalize $1$ to $k$, and consider the expression $$\frac{(n+k)^n}{n^{n+1}}=\left(\frac{n+k}{n}\right)^n\frac{1}{n}=\left(1+\frac{k}{n}\right)^n\frac{1}{n}$$ The first term, i.e. $\left(1+\frac{k}{n}\right)^n$, is an expression that is strictly increasing with $n$, and the limit is $e^k$. Hence the expression will be less than 1 whenever $n>e^k$. In the OP, $k=1$, so the expression will be less than 1 whenever $n>e$, i.e. $n\ge 3$.

A proof without induction :

Consider the function $f(x)=x^{\frac{1}{x}}$ .

The problem is equivalent with $f(n)>f(n+1)$ .

It's easy to find the derivative of $f$ : $$f'(x)=x^{\frac{1}{x}-2}(1-\ln x)$$ so $f$ is strictly decreasing for every $x>e$ and so the conclusion holds (because $e<3$) :

$$f(n)>f(n+1)$$ which means that :

$$n^{(n+1)}>(n+1)^n$$

• Please explain how the problem is equivalent to $f(n)>f(n+1)$. – vadim123 Sep 6 '15 at 15:46
• because $n^{n+1}>(n+1)^n$ is equivalent with $n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$ (just raise both sides to the power $\frac{1}{n(n+1)}$ to see this ) which is the same as $f(n)>f(n+1)$ – user252450 Sep 6 '15 at 15:48

The inequality at stake is equivalent to $n > \left( 1+\frac{1}{n}\right)^n$.

Taking log yields $\frac{\ln n}{n} >\ln \left(1+\frac{1}{n} \right)$.

This is true since the concavity of $\log$ implies the stronger $\frac{1}{n} >\ln \left(1+\frac{1}{n} \right)$.

Using this argument, your inequality can be refined to $(n+1)^n<e^1n^n$

That is equivalent to proving that $a_n = \left(1+\frac{1}{n}\right)^n$ is an increasing sequence.

That follows, in a very straightforward way, from the AM-GM inequality, since:

$$\left(1+\frac{1}{n}\right)^n\cdot 1\leq\left(\frac{1+n\cdot\left(1+\frac{1}{n}\right)}{n+1}\right)^{n+1}=\left(1+\frac{1}{n+1}\right)^{n+1}.$$