Proof: $\sqrt[n]{n} > \sqrt[n+1]{n+1}$

Question

How can I prove that $\sqrt[n]{n} > \sqrt[n+1]{n+1}$ for $n \in \mathbb{N} \setminus \{ 1,2 \}$ ?

My approach:

Step 1:

$n_0 := 3 \qquad \sqrt[3]{3} > \sqrt[4]{4}$

which is true.

Step 2:

$\sqrt[n]{n} > \sqrt[n+1]{n+1}$ is true for any $n \in \mathbb{N} \setminus \{ 1,2 \}$

Step 3:

$A(n) \rightarrow A(n+1)$ for any $n \in \mathbb{N}_{>3}$

$\sqrt[n+1]{n+1} > ... > A(n+1)$

And here i don't know how to proceed.

Can someone give me a tip?

Answer 1

$\sqrt[n]{n} > \sqrt[n+1]{n+1}$

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

$$\begin{array}{l}\Leftrightarrow n^{n+1}>n^n+n\times n^{n-1}+\frac{n(n-1)}{2}n^{n-2}+...+1\\\end{array}$$

From there you shouldn't have much trouble.

Answer 2

Step 1. $f(x)=\frac{\log x}{x}$ is a decreasing function over $(e,+\infty)$, since over such set: $$ f'(x) = \frac{1-\log x}{x^2} < 0.$$ Step 2. Our inequality is equivalent to: $$ \forall n\geq 3,\quad f(n+1)<f(n).$$

Answer 3

Let $f(x)=x^{1/x}$ then $f^{\prime}(x)=f(x)\frac{1-\log x}{x^2}<0$ for $x>e.$

Therefore, $f(x)$ is decreasing function for $x>e.$ Thus $f(n)>f(n+1)$ for $n\geq 3.$

Answer 4

$\sqrt[n]{n} > \sqrt[n+1]{n+1} \iff n^{n+1} > (n+1)^n \iff n > \left(1+\dfrac1n\right)^n$

Now $ 2 \le \left(1+\dfrac1n\right)^n < 3$ for all $n$.

Answer 5

Raising to the power $n(n+1)$ this is equivalent to $$ n^{n+1}>(n+1)^n $$ or $$ n>\frac{(n+1)^n}{n^n}=\left(1+\frac{1}{n}\right)^{\!n} $$ Assuming you know that $$ \left(1+\frac{1}{n}\right)^{\!n}<e<3 $$ you're done. See How is $a_n=(1+1/n)^n$ monotonically increasing and bounded by $3$?