$\lim\limits_{n\to \infty}\sup (1/n)^{1/n}=1$ I am quite stuck as how to how I can prove this. My attempt so far is to show that for any $x\in (0,1)$ we can find an , n such that $x^n<1/n$. But I got stuck there playing with the derivative of $a^y-1/y$.
 A: \begin{align*}
\left(\frac{1}{n}\right)^\frac{1}{n} &= e^{\ln\left(\left(\frac{1}{n}\right)^\frac{1}{n}\right)}\\
&= e^{\frac{\ln\left(\frac{1}{n}\right)}{n}}\\
&= e^{-\frac{\ln(n)}{n}}
\end{align*}
Note that by L'Hospital's Rule, $\lim\limits_{n\to\infty} \frac{\ln(n)}{n} = 0.$
So 
\begin{align*}
\lim_{n\to\infty} \left(\frac{1}{n}\right)^\frac{1}{n} &= \lim_{n\to\infty} e^{-\frac{\ln(n)}{n}}\\
&= e^{-\lim_{n\to\infty} \frac{\ln(n)}{n}}\\
&= e^0\\
&= 1.
\end{align*}
Since the limit exists, the lim sup is equal to it.
A: One has
$$\log{\left({1\over n}\right)^{1\over n}}={1\over n}\log{1\over n}$$
Now $\lim\limits_{x\to 0}x\log{x}=0$ so the limit of the sequence above exists and is $1$. And so the $\limsup$ is $1$ as well
A: non standard solution
Consider the series $\sum_{n \in \mathbb N} \frac1n \cdot z^n$. Since $\sum_{n \in \mathbb N} z^n$ converges for all $z \in \mathbb C$ with $|z| \lt 1$ and it holds that $|\frac 1n \cdot z| \le |z|$, the series also converges for these $z$. Furthermore, $\sum_{n \in \mathbb N} \frac1n \cdot 1$ does not converge, ie. the the convergence radius is $1$.
By the convergence radius formula one gets $1 = \frac{1}{\limsup_{n \rightarrow \infty} |\frac1n|^\frac1n}$.
