Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
Do you know that $n^{1/n}$ has limit $1$? –  David Mitra Sep 13 '12 at 14:46
3  
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
1  
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

4 Answers 4

up vote 7 down vote accepted

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

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

share|improve this answer
    
@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 $n\to \infty$.

share|improve this answer

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

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.