Does $\alpha_n = \left(\frac{1}{n}\right)^{\frac{1}{n}}$ converge to $1$? Does the sequence $\alpha_n = \left(\frac{1}{n}\right)^{\frac{1}{n}}$ converge to $1$? If yes, how can i show that? I tried various simple methods unsuccessfully. 
 A: $$\underbrace{\lim_{n \to \infty} \left( \dfrac1n \right)^{1/n} = \lim_{x \to 0^+} x^x}_{1/n = x} = \underbrace{\lim_{x \to 0^+} \exp(x \log x) = \exp \left( \lim_{x \to 0^+} x \log x\right)}_{\exp(y) \text{ is continuous}} = \exp(0) = 1$$
To show $\lim_{x \to 0^+} x \log x = 0$, take $x = \exp(-t)$, then we have $$0 \geq \lim_{x \to 0^+} x \log x = \lim_{t \to \infty} \exp(-t) \times(-t) = - \lim_{t \to \infty} \dfrac{t}{\exp(t)} \geq - \lim_{t \to \infty} \dfrac{t}{1+t+t^2/2} = 0$$
A: Here is another proof:
Since $\displaystyle \bigg(1+\frac{2}{\sqrt{n}}\bigg)^n\geq 1 +\frac{n(n+1)}{2}\times\bigg(\frac{2}{\sqrt{n}}\bigg)^2>n$, we have
　　　　$\displaystyle 1\leq n^{1/n} \leq 1+\frac{2}{\sqrt{n}}$
Now the conclusion follows by letting $n\to \infty$.
A: Since
$$ \left( \frac{1}{n} \right)^{\frac{1}{n}} = e^{-\frac{\ln(n)}{n}}$$
and $\exp$ is continuous you only have to show that
$$ \underset{n \rightarrow \infty}{\lim}  \frac{\ln(n)}{n} = 0 $$
This can be achieved easily noticing that $\ln(x)\le x-1$ for all $x\ge 1$. Then for $n \ge 1$
$$ 0 \le \frac{\ln(n)}{n} \le \frac{2\ln(\sqrt{n})}{n} \le \frac{2(\sqrt{n}-1)}{n} \underset{n \rightarrow \infty}{\longrightarrow} 0$$
A: Use Cauchy-d'Alembert criterion for the denominator and you're done.
Chris.
