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I wanted to find a way of doing this without employing the exponential. So, let $a_n = n^{\frac{1}{n}} -1.$ Then \begin{align*} n &= (a_n +1)^n \\ &= \sum_{k=0}^n \binom{n}{k} a_n^{k} \\ &\geq \frac{n(n-1)}{2} a_n^2 \\ &\geq 0. \end{align*} So that $$0\leq a_n \leq \sqrt{\frac{2}{n-1}}.$$ Letting $n$ sail to infinity we see that $$\lim a_n =0$$ which is what we wished to prove.

My question is this: I assumed $a_n \geq 0$. Is this true? How would a proof of this look? Any hints are appreciated.

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  • $\begingroup$ Just another idea complementary to the answers below: you may write $n^{\frac{1}{n}} = \exp(\frac{\log n}{n})$. As $\frac{\log n}{n}$ goes to $0$ as $n \rightarrow \infty$ by taking positive values (you may apply l'Hopital), you can conclude. $\endgroup$ – bgsk Nov 22 '19 at 7:19
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If $n^{1/n} < 1$, then $n <1^n = 1$ (contradiction).

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    $\begingroup$ I object to calling this a hint. This is a full proof, albeit a brief one. $\endgroup$ – Arthur Nov 22 '19 at 6:50
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    $\begingroup$ @Arthur: Objection duly noted. Would you like me to remove “Hint” $\endgroup$ – RRL Nov 22 '19 at 6:54
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    $\begingroup$ Personally, yes, I would. But ultimately it's up to you. I'm just a random person on the internet, and you are entirely allowed to disregard my opinion on what the word "hint" ought to mean. Also, it's not like it's a terribly important thing. $\endgroup$ – Arthur Nov 22 '19 at 6:58
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Or you can use a simple fact that $ x \ge y\ge 0, p > 0 \Rightarrow x^p \ge y^p$ since $n \ge 1$ and $\frac{1}{n} > 0$.

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While it's true that $a_n \ge 0$ for each $n$ (as RRL has explained), it's not necessary to use that in order to prove the result. Let $0 < \epsilon < 1$, and choose $N > \frac{2}{\epsilon^2}$. If $n > N$, then $$(1 + \epsilon)^n > 1 + \frac{n(n-1)}{2}\epsilon^2 > 1 + (n-1) = n > 1 > (1 - \epsilon)^n$$The inequalities imply $|a_n| < \epsilon$ for all $n > N$. Thus $\lim\limits_{n\to \infty} a_n = 0$.

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  • $\begingroup$ No. His derivation assumed $a_n \ge 0$ to get the inequality you mentioned(so that all terms in summation are nonnegative). I am not saying it can not be fixed. There are many ways to prove. $\endgroup$ – Xiaohai Zhang Nov 22 '19 at 10:54
  • $\begingroup$ @XiaohaiZhang note that I have absolute value bars around $a_n$ while the OP does not. My inequality with $a_n$ does not assume the $a_n$ are nonnegative. $\endgroup$ – kobe Nov 22 '19 at 14:27
  • $\begingroup$ Proving $n \ge \frac{n(n-1)}{2}a_n^2$ requires the assumption of $a_n \ge 0$. $\endgroup$ – Xiaohai Zhang Nov 22 '19 at 16:42
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I would have done it this way, pretty much the same as you: $$n^{\frac{1}{n}}>1\text{ for } n>1\text{ therefore } n^{\frac{1}{n}}=1+a_n\text{ for some $a_n>0$}$$

Thus $$(n^{\frac{1}{n}})^n=n=(1+a_n)^n$$By the Binomial Theorem: $$n=(1+a_n)^n=1+na_n+\frac{1}{2}n(n-1)a_n^2+...\geq1+\frac{1}{2}n(n-1)a_n^2$$ therefore $$n-1\geq\frac{1}{2}n(n-1)a_n^2\iff\frac{2}{n}\geq a_n^2\iff a_n\leq\sqrt{\frac{2}{n}}$$

By the Archimedean Property, there exists a natural number $K$ such that $$\frac{2}{K}<\epsilon^2\iff\sqrt{\frac{2}{K}}<\epsilon$$ for all $\epsilon>0$. Hence if we take $n\geq\sup\{2,K\}$ (where the $2$ comes from the inequality $n^{\frac{1}{n}}>1$ for $n>1$) , then we have $$0<|n^{\frac{1}{n}}-1|=|a_n|\leq\sqrt{\frac{2}{n}}<\epsilon\space\space\space ,\forall\epsilon>0$$

Hence $\lim n^{\frac{1}{n}}=1$.

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