Show $n^{\frac{1}{n}}$ is decreasing for $n \ge 3$

Question

How to show $\displaystyle n^{\frac{1}{n}}$ is decreasing for $n \ge 3$ ?

Answer 1

$$(n+1)^{\frac{1}{n+1}} \leq n^{\frac{1}{n}}$$

$$\iff (n+1)^{n} \leq n^{n+1}$$

$$\iff (1+\frac{1}{n})^{n} \leq n$$ and $$(1+\frac{1}{n})^{n} = \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{1}{n^k} \leq \sum_{k=0}^n \frac{1}{k!} \leq e < 3$$

Answer 2

$n^{1/n}=e^{\frac{1}n\ln(n)}$ and $\exp$ is increasing hence you only need to prove that $\frac{\ln(x)}x$ is decreasing for $x\ge3$, which should be easy :)

(if you still can't do it, tell me)

Answer 3

Let $f(x) = x^{\frac{1}{x}} \to \ln f(x) = \dfrac{\ln x}{x} \to \dfrac{f'(x)}{f(x)} = \dfrac{1-\ln^2 x}{x^2} < 0$ when $x > 3$. Since $n > 3$, $f'(n) < 0$, and this means $n^{\frac{1}{n}}$ decreases.

Answer 4

$\dfrac{d}{dn}[n^{\frac{1}{n}}]$

$=\dfrac{d}{dn}[e^{\frac{log(n)}{n}}]$

$=n^{1/n-2}(1-log(n))$ We want to prove that for all $n \ge 3, \frac{d}{dn}[f(n)]<0$.

For all $n>e, f'(n)< 0 $ because $(1-log(n))<0$, and all exponentials are positive.

Clearly $3>e$, so this statement is always true.

Answer 5

Late answer but $n>e$ so $$n^{1/n} > e^{1/n}>1 + 1/n = \frac{n+1}{n} \implies n^{1+1/n}>n+1 \implies n^{1/n} > (n+1)^{1/n+1}$$