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Suppose $x>0$ is a real number. I have to prove that $\lim_{n\rightarrow{\infty}}x^{1/n}=1$, and I'm working in the context of sequences, not continuous functions. Here's what I've done so far: there are two cases, $x\geq{1}$ and $0<x<1$. In the first case, it is possible to show that $(x^{1/n})_{n=1}^{\infty}$ is a decreasing sequence, and moreover is bounded from below by $1$, so we know it converges to a real number and furthermore, that: $$\lim_{n\rightarrow{\infty}}x^{1/n}=\inf(x^{1/n})_{n=1}^{\infty}$$ The problem then becomes to verify that the infimum of the sequence is indeed $1$. We already know that $1$ is a lower bound, so the task becomes to show that $1$ is the greatest lower bound. I have tried to employ a proof by contradiction of assuming we could find some lower bound $L$ such that $L>1$, but I'm not allowed to use logarithms so I'm stuck at this point (a similar argument to the above would establish for the case $0<x<1$ that $\lim_{n\rightarrow{\infty}}x^{1/n}=\sup(x^{1/n})_{n=1}^{\infty}$, etc, but I'm stuck for the same reason there). Any suggestions on how to move forward would be appreciated (in case you're curious, this is Exercise 6.5.3 in Tao's Analysis 1). Note there are some duplicate questions, but all of the methods which were used to prove them are "too advanced" for my purposes. It needs be as elementary as possible, appealing to the most basic properties of suprema, infima, and possibly limit superior and limit inferior, etc.

SurfaceIntegral
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  • It is a similar question, for the case $0<x<1$, but the methods that were used to prove it are far too advanced for my context. I need a much more elementary proof. – SurfaceIntegral Dec 23 '20 at 15:31
  • For $0<x<1$: If $L<1$ then $L^2<L<1$ and since $x^{1/n}\leq L$ for all $n$ we would also have $x^{1/n}=x^{2/(2n)}\leq L^2<L$ for all $n$, contradicting $L=\sup_n x^{1/n}$. For $x>1$ it is similar with $1<\sqrt{L}<L$ instead. – user10354138 Dec 23 '20 at 15:38
  • Also for $x > 1$ we have $x = [1 + (x^{1/n} - 1)]^n > 1 + n(x^{1/n}-1)$ by Bernoulli's inequality which implies that $0 < x^{1/n}-1 < (x-1)/n \to 0$. Does Bernoulli's inequality qualify as elementary? The case $0 < x < 1$ can be handled similarly with $x^{1/n} = 1/(1 +a)$ where $a > 0$. – RRL Dec 23 '20 at 15:48
  • No, not elementary enough. – SurfaceIntegral Dec 23 '20 at 15:48
  • $0\leq L=\sup_n x^{1/n}$, so $x^{2/(2n)}=x^{1/(2n)} x^{1/(2n)}$ and both factors are $<L$. So $x^{1/n}\leq L^2$ for all $n$ and hence $\sup_n x^{1/n}$ is bounded above by $L^2$ which is smaller than $L$, contradicting $L$ being the least upper bound. – user10354138 Dec 23 '20 at 15:51
  • Let $n>m$ be integers and let $x>1$ be a real number. Then $0<{1/n}<{1/m}$, and I had elsewhere as a lemma that if $x>1$ and $q,r$ are rationals, then $x^{q}>x^{r}$ if and only if $q>r$. – SurfaceIntegral Dec 23 '20 at 16:07
  • @RRL The reason I'm saying Bernoulli's inequality is "not elementary enough" is that I just don't think Tao would expect a reader who's never seen it before to have it as some sort of sudden insight, and nor has it been introduced in the book thus far. – SurfaceIntegral Dec 23 '20 at 16:09
  • @user10354138: just to be clear: your comment assumes $x<1$, right? – TonyK Dec 23 '20 at 16:40

1 Answers1

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Suppose $x^{1/n}$ tends to some $L>1$. Then $L^2>L$, so there exists $N$ such that $x^{1/n}<L^2$ whenever $n\ge N$. But $x^\frac{1}{2n}=\sqrt{x^{1/n}}$, which is less than $L$ if $n\ge N$. This contradicts the assumption that $x^{1/n}$ tends monotonically to $L$ from above.

And for the case $0<x<1$, just apply the above to $1/x$.

TonyK
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  • Your solution is pretty good, but I have to confess I liked user10354138's solution a little bit better! Nonetheless, I'll accept it as the best answer. – SurfaceIntegral Dec 23 '20 at 16:34