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Given that $a_n>0, \forall{n}, \ \lim_{n \to \infty}\ a_n =a >0$, prove that$$ \ \lim_{n \to \infty}\ \sqrt[n]{a_n} =1 $$

This what I did: Let $a_n=\frac{1}{n}$ then $$\lim_{n \to \infty}\ \sqrt[n]{a_n} = \lim_{n \to \infty} \left(\frac{1}{n}\right)^\frac{1}{n} =\lim_{n \to \infty} \frac{1}{n^\frac{1}{n}} =1 $$

But I don't know whether this right. Anyone can suggest an alternative way please.

Martin Sleziak
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Ramos
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  • Which limit exists? – Julián Aguirre Sep 26 '15 at 08:54
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    Your argument has nothing to do with the question. You attempt to prove it for only one sequence, which wouldn't be a proof even if the sequence were one of the ones being considered. However, the limit of your sequence is $0$, so it has no bearing on what is being proved in any case. – Matt Samuel Sep 26 '15 at 09:06
  • @JuliánAguirre my apologies. i forgot to specify the limit. – Ramos Sep 26 '15 at 09:16
  • This might be helpful: http://math.stackexchange.com/questions/69386/inequality-involving-limsup-and-liminf and http://math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n Probably also other questions linked to 69386 and to 76743. – Martin Sleziak Sep 26 '15 at 14:44

2 Answers2

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To prove it by that way is not enough. $a_n=\frac{1}{n}$ is but a special case, even if you show it works on $\frac{1}{n}$, the other possible $a_n$ is still undefined.

Since $\lim \limits_{n \to \infty} a_n=a>0$, We take $\epsilon_0=\frac{1}{2} a>0$, then there is $N_0>0$, when $n>N_0$, $$\frac{1}{2} a<a_n<\frac{3}{2}a.$$ Thus, $$\sqrt[n]{\frac{1}{2} a}<\sqrt[n]{a_n}<\sqrt[n]{\frac{3}{2}a}.$$

(You can see the choice of $\epsilon_0$ is quite arbitrary.)

By squeeze theorem you can get the result.

P. S. Proposition: For any $c>0$, $\lim \limits_{n \to \infty} \sqrt[n] c=1$.

Proof: For $\epsilon>0$, assume that $c>1$. $$\begin{align} |\sqrt[n] c-1|<\epsilon &\Leftrightarrow \sqrt[n] c < 1+\epsilon \\ &\Leftrightarrow \frac{1}{n} \ln c < \ln (1+\epsilon) \\ &\Leftrightarrow n>\frac{\ln c}{\ln (1+\epsilon)} \end{align}$$ Hence, take $N=\lceil \frac{\ln c}{\ln (1+\epsilon)} \rceil$, when $n>N$, $|\sqrt[n] c-1|<\epsilon$.

For $0<c<1$, the trick is similar. If $\epsilon >1$, $|\sqrt[n] c-1|<\epsilon$ always holds.

If $0<\epsilon<1$, notice that $\ln c<0$, $\ln (1-\epsilon)<0$. $$\begin{align} |\sqrt[n] c-1|<\epsilon &\Leftrightarrow 1-\epsilon < \sqrt[n] c \\ &\Leftrightarrow \ln (1-\epsilon) < \frac{1}{n} \ln c\\ &\Leftrightarrow n>\frac{\ln c}{\ln (1-\epsilon)} \end{align}$$ Hence, take $N=\lceil \frac{\ln c}{\ln (1-\epsilon)} \rceil$, when $n>N$, $|\sqrt[n] c-1|<\epsilon$.

From above we know $$\lim \limits_{n \to \infty} \sqrt[n] c=1.$$

Asydot
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Suppose $a>1$. Let $b_n=\sqrt[n]{a_n}-1$. If we prove $b_n \rightarrow 0$, we are done.

We know that:

$$1+nb_n \leq (1+b_n)^n=a_n.$$

This implies:

$$b_n \leq \frac{ a_n-1}{n}.$$

Since we are supposing $a>1$, we also have that (at least for sufficiently large $n$):

$$0 \leq b_n.$$

Combining both inequalities and using the squeeze theorem yields the result.

In the case that $a<1$, apply the previous result with the sequence $\sqrt[n]{\frac{1}{a_n}}$.

Aloizio Macedo
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  • If $a < 1$, let $a = 1/(1+b)$. Then $(1+b)^n > 1+nb > nb$ so $a^n < 1/(nb)$. All this is in "What is Mathematics" by Courant and Robbins. Similarly, $(1+n^{-1/2})^n > 1+n^{1/2} > n^{1/2}$ so $n^{1/n} < (1+n^{-1/2})^2 < 1+3n^{-1/2} \to 1$. – marty cohen Sep 27 '15 at 04:29