$ \{ n^{1/n} \}_{ n \in \mathbb{N}} $ How can I prove that $ \{ n^{1/n} \}_{ n \in \mathbb{N}} $
is a decreasing sequence after some $n_0 \in \mathbb{N} $;
and also find out such $n_0$?
Thanks in advance.
 A: New attempt
$$
\left(\frac{n^{1/n}}{(n+1)^{1/(n+1)}}\right)^{n\cdot(n+1)} = \frac{n^{{n+1}}}{(n+1)^n} = n\left(\frac{n}{n+1}\right)^n = n\left(1-\frac{1}{n}\right)^n
$$
whhere the last $\left(1-\frac{1}{n}\right)^n$ is closing in on $\frac{1}{3}$ from below (from $n=4$, it is bigger than $0.3$), so from that point on, the whole expression will be bigger than $1$, and all subsequent terms of the sequence will be smaller than their predecessor.
A: $$\sqrt[n+1]{n+1} \leq \sqrt[n]{n} \Leftrightarrow (n+1)^n \leq n^{n+1} \Leftrightarrow (1+\frac{1}{n})^n \leq n$$
The LHS is increasing to $e$, thus the inequality holds for sure for $n \geq 3$., and it is easy to see that 
$$(1+\frac{1}{2})^2 > 2$$
Thus, $n_0=3$.
A: Let $f(x) = x^\frac{1}{x}$. Differentiate this function and you will be able to spot a value of $x_0$ after which $f'(x) < 0$. Let $n_0$ be the first integer after this, and then you'll have $f$ decreasing on $[n_0,\infty)$.
A: Since Un the term of your series is positive you can consider Un+1/Un.
You have Un+1/Un = exp[(n ln(n+1) - (n+1) ln(n))/(n (n+1))] = exp[(ln(1+1/n) - ln(n))/n(n+1)].
As you know ln(1+1/n) <= 1/n. Hence, Un+1/Un <= exp[(1 - ln (n))/(n(n+1))]. So after n>=3, the expression in the exponential is negative and Un+1/Un is less than 1.
Hope this helps.
