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For the sequence $a_n=n^{\frac 1n}$, first I have shown that the sequence is decreasing for $n \ge 3$ and bounded below by $1$. So it must converge and I call its limit $\lim\limits_{n \to \infty} a_n=a$. Its subsequence $a_{2n}$ must also converge to this limit, so $\lim\limits_{n \to \infty} a_{2n}=\lim\limits_{n \to \infty} a_n=a$.

Then I write $a_{2n}$ as $$a_{2n}=(2n)^{\frac 1 {2n}} = 2^{\frac 1 {2n}} \cdot (n^{\frac 1 {n}})^{\frac 1 {2}} = 2^{\frac 1 {2n}} \cdot \sqrt a_n $$

And when I take the limit as $n$ goes to infinity, I get the following

$$\lim\limits_{n \to \infty} a_{2n}=a= \lim\limits_{n \to \infty} (2^{\frac 1 {2n}} \cdot \sqrt a_n) = 2^0 \cdot \sqrt a$$

since $\lim\limits_{n \to \infty} \frac 1 n=0$ and $\lim\limits_{n \to \infty} a_n =a$.

From here I simply solve $a=\sqrt a$, get $a=1$ and conclude that the limit is $1$.

I am not sure if my proof is correct because in the following links, I couldn't find any solution similar to this:

Prove sequence $a_n=n^{1/n}$ is convergent

Proof of $\lim_{n\to \infty} \sqrt[n]{n}=1$

Can you please confirm my solution?

user666150
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    This seems correct (except that $-1$ is not at all a solution of $a=\sqrt{a}$). However, let me say that this is far more complicated than writing $n^{1/n}=\exp(\ln(n)/n)$ and seing directly that this converges to $1$ ! – TheSilverDoe Oct 21 '20 at 10:50
  • It depends on what you mean by complicate, @TheSilverDoe Defining $\exp$ and $\ln$ are indeed complicate. – Arctic Char Oct 21 '20 at 10:54
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    @TheSilverDoe: the matter of "being complicated" is subtle, especially if one includes the theoretical material the solution is based on; in this measure, this solution is among the best. – metamorphy Oct 21 '20 at 10:54
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    @TheSilverDoe: Ah yes, for some reason I have solved $a^2=a$ instead of $a=\sqrt a$. – user666150 Oct 21 '20 at 10:55
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    Show that the limit of $\ln a_n=\frac{\ln n}{n}$ is zero – Raffaele Oct 21 '20 at 11:48
  • I think it's okay. In an earlier edition you claimed $-1$ was a solution to $a^2 = a$ but $-1 < 1$. FWIW $0$ is also a solution (and a solution to $a = \sqrt a$) but also less than the lower bound of $1$. ....for what its worth.... which is probably not much. – fleablood Oct 21 '20 at 12:25
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    Please write $\cdot$ (\cdot), or just nothing, instead of $*$. – sicmath Oct 21 '20 at 14:21

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