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Possible Duplicate:
$\lim_{n \to +\infty} n^{\frac{1}{n}} $
Limit of the sequence $\lim_{n\rightarrow\infty}\sqrt[n]n$

I want to prove that $\lim\limits_{n\rightarrow\infty}\sqrt[n]{n}=1$

Proof: We can write $\sqrt[n]{n}=1+a_n$ and so $n=(1+a_n)^n$ with $a_n\geq0$.

So by the binomial theorem it's $n=(1+a_n)^n=\sum\limits_{k=0}^{n}{n\choose k}a_n^k>{n\choose2}a_n^2$ for all $n\geq2$ $(\ast)$

Now it's $n>{n\choose 2}a_n^2$ what is equivalent to $\frac{2}{n-1}>a_n^2$ what is $\geq0$. By taking the limit for $n\rightarrow\infty$ its $0>\lim\limits_{n\rightarrow\infty}a_n\geq0$, a contradiction.

So my question is just about the '$>$' and if this is a mistake or can you do it like this?

Appendix: Of course I could take in $(\ast)$ $n\geq1+{n\choose2}a_n^2$ and so I will get $\lim\limits_{n\rightarrow\infty}\sqrt\frac{2}{n}=0\geq\lim\limits_{n\rightarrow\infty}a_n\geq0$ and so $\lim\limits_{n\rightarrow\infty}a_n=0$ and so $\lim\limits_{n\rightarrow\infty}\sqrt[n]{n}=1+0=1$

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  • $\begingroup$ But your (*) requires that $a_n$ be $>1$, right? How can one assure this? I know that is true, but your way of writing it seems misleading... Thanks for the explanation. $\endgroup$
    – awllower
    Dec 9, 2012 at 19:24

3 Answers 3

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Let $A=n^{\frac1n}$

So, $\log A=\frac{\log n}n$

$\lim_{n\to \infty}\log A=\lim_{n\to \infty}\frac{\log n}n$ which is of the form $\frac{\infty}{\infty}$

So, applying L'Hospital's Rule, $\lim_{n\to \infty}\log A=\lim_{n\to \infty}\frac{\log n}n=\lim_{n\to \infty}\frac1n=0$

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    $\begingroup$ sry but this is not what I've wanted. $\endgroup$
    – user45389
    Dec 9, 2012 at 18:01
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Your proof is correct. Indeed, $$0<a_n<\sqrt{\frac{2}{n-1}}$$ and by the squeeze theorem, $a_n\to 0$. This proves $n^{1/n}\to 1+0=1$. There is no "contradiction" however.

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  • $\begingroup$ but isn't the squeeze theorem just using $0\leq a_n\leq \sqrt{\frac2{n-1}}$? $\endgroup$
    – user45389
    Dec 9, 2012 at 18:14
  • $\begingroup$ Strict inequalities imply non strict inequalities... $\endgroup$
    – Nameless
    Dec 9, 2012 at 18:15
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Yes, $(*)$ is right, because $a_n \geq 0$.

You can also get $(*)$ the following way:

If $n$ is even:

$$n=(1+a_n)^n=[(1+a_n)^2]^{\frac{n}{2}}\geq (1+a_n^2)^{\frac{n}{2}} $$

Now use Bernoulli. For $n$ odd, you can do the same but with

$$n=(1+a_n)^n>[(1+a_n)^2]^{\frac{n-1}{2}}$$

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