Show $n^{\frac{1}{n}}$ is decreasing for $n \ge 3$ How to show $\displaystyle n^{\frac{1}{n}}$ is decreasing for $n \ge 3$ ?
 A: $$(n+1)^{\frac{1}{n+1}} \leq n^{\frac{1}{n}}$$
$$\iff (n+1)^{n} \leq n^{n+1}$$
$$\iff (1+\frac{1}{n})^{n} \leq n$$
and
$$(1+\frac{1}{n})^{n} = \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{1}{n^k} \leq \sum_{k=0}^n \frac{1}{k!} \leq e < 3$$
A: $n^{1/n}=e^{\frac{1}n\ln(n)}$ 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)
A: 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.
A: $\dfrac{d}{dn}[n^{\frac{1}{n}}]$
$=\dfrac{d}{dn}[e^{\frac{log(n)}{n}}]$
$=n^{1/n-2}(1-log(n))$
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.
A: Late answer but $n>e$ so $$n^{1/n} > e^{1/n}>1 + 1/n = \frac{n+1}{n} \implies n^{1+1/n}>n+1 \implies n^{1/n} > (n+1)^{1/n+1}$$
