# 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.

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Do you know that $n^{1/n}$ has limit $1$? – David Mitra Sep 13 '12 at 14:46
First show that $\frac{\log n}{n}\to 0$ as $n\to \infty$. Then take $e^{-\frac{log n}{n}}\to e^{-0}$ – Thomas Andrews Sep 13 '12 at 14:48
If you don't want to remark that $(1/n)^{1/n}=1/(n^{1/n})$, you may recall that $\lim_{x \to 0+} x^x=1$. – Siminore Sep 13 '12 at 14:51

$$\underbrace{\lim_{n \to \infty} \left( \dfrac1n \right)^{1/n} = \lim_{x \to 0^+} x^x}_{1/n = x} = \underbrace{\lim_{x \to 0^+} \exp(x \log x) = \exp \left( \lim_{x \to 0^+} x \log x\right)}_{\exp(y) \text{ is continuous}} = \exp(0) = 1$$

To show $\lim_{x \to 0^+} x \log x = 0$, take $x = \exp(-t)$, then we have $$0 \geq \lim_{x \to 0^+} x \log x = \lim_{t \to \infty} \exp(-t) \times(-t) = - \lim_{t \to \infty} \dfrac{t}{\exp(t)} \geq - \lim_{t \to \infty} \dfrac{t}{1+t+t^2/2} = 0$$

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@DavidMitra: I think it's fair to say that the ${}^+$ is at best redundant when no other limit is possible. – Ben Millwood Sep 13 '12 at 14:57
@DavidMitra Thanks. corrected. – user17762 Sep 13 '12 at 14:57

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 $n\to \infty$.

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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 $n \ge 1$ $$0 \le \frac{\ln(n)}{n} \le \frac{2\ln(\sqrt{n})}{n} \le \frac{2(\sqrt{n}-1)}{n} \underset{n \rightarrow \infty}{\longrightarrow} 0$$

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To show that $\ln(x)\le x-1$ for all $x\ge 1$ just study the function $\ln(x)-x+1$ and show it is always negative ;) – vanna Sep 13 '12 at 15:33

Use Cauchy-d'Alembert criterion for the denominator and you're done.

Chris.

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