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I know that

$$\lim_{n\to\infty} \sqrt[n]{ \sqrt[n]{n} - 1 } = 1,$$

but I'm unable to prove it. I could easily estimate that it's at most $1$, but my best estimation from below is that the limit is greater than $0$.

Doing this from the definition doesn't lead me anywhere either.

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    $\begingroup$ Have you tried appealing to l'Hopital's rule at all? $\endgroup$ – πr8 Jan 22 '16 at 11:44
  • $\begingroup$ Seems an overkill to use l'Hospital or Taylor for seqs ;) $\endgroup$ – marmistrz Jan 22 '16 at 15:37
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The numbers $n^{k/n}$ with $0\leq k\leq n-1$ are all between $1$ and $n$. By the formula for the sum of finite geometric series it follows that $$n\leq\sum_{k=0}^{n-1} n^{k/n}={n-1\over n^{1/n}-1}\leq n^2\qquad(n\geq2)\ .$$ From this we infer $${1\over 2n}<{n-1\over n^2}\leq n^{1/n}-1\leq{n-1\over n}<1\ .$$ Using $\lim_{n\to\infty} (2n)^{1/n}=1$ and the squeeze theorem one then concludes that $$\lim_{n\to\infty}\bigl(n^{1/n}-1)^{1/n}=1\ .$$

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    $\begingroup$ Great answer but would the Op understand this? $\endgroup$ – Arbuja Jan 22 '16 at 13:43
  • $\begingroup$ I'll have to analyze it on my own, but seems legit at the first sight ;) $\endgroup$ – marmistrz Jan 22 '16 at 15:34
  • $\begingroup$ I think you should justify the non-obvious ineq $n\leq\sum_{k=0}^{n-1} n^{k/n}$ $\endgroup$ – marmistrz Jan 23 '16 at 9:13
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    $\begingroup$ As $0\leq k/n\leq 1$ each term in the sum is $\geq1$. $\endgroup$ – Christian Blatter Jan 23 '16 at 9:18
  • $\begingroup$ A beautiful proof ;) $\endgroup$ – marmistrz Jan 23 '16 at 11:24
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Since deriving stuff like $f(x)^{g(x)}$ is a nightmare (at least to me), always try to apply the logarithm in these cases, and see if you can come up with something easier.

$$\lim_{n \to \infty} (n^{\frac{1}{n}} -1)^{\frac{1}{n}}$$

$$\lim_{n \to \infty} \log(n^{\frac{1}{n}} -1)^{\frac{1}{n}} = \lim_{n\to \infty} \frac{\log (n^{\frac{1}{n}} -1)}{n}$$ $$ = \lim_{n\to \infty} \frac{\log (n^{\frac{1}{n}} -1)}{n}$$

Now since $\log (n^{\frac{1}{n}} -1) \leq (n^{\frac{1}{n}} -1) \leq n^{\frac{1}{n}}$ we get $$\lim_{n\to \infty} \frac{\log (n^{\frac{1}{n}} -1)}{n} \leq \lim_{n \to \infty} \frac{1}{n^{\frac{n-1}{n}}} = 0$$

Which means that the above limit is $1$, since $\log(1)=0$.

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  • $\begingroup$ One also has to argue that it isn't $<0$... $\endgroup$ – punctured dusk Jan 22 '16 at 11:53
  • $\begingroup$ It's pretty easy. Moreover, the OP did that $\endgroup$ – AnalysisStudent0414 Jan 22 '16 at 11:54
  • $\begingroup$ No, the OP did not do that, but it's easy, because $n^{1/n}>1$ for all $n$. $\endgroup$ – yo' Jan 22 '16 at 13:28
  • $\begingroup$ @AnalysisStudent0414 the OP showed that the original limit is $\geq 0$, not the one after taking $\log$. So we would need $\log(n^{1/n}-1)>0$, which is not true, hence we need something clever such at l'Hopital or a non-trivial lower bound for $n^{1/n}-1$, as in Christian Blatter's answer. $\endgroup$ – punctured dusk Jan 22 '16 at 14:26
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    $\begingroup$ You simply showed that the limit is $\le 1$. But you still need to justify that it's not $\ge 1$ $\endgroup$ – marmistrz Jan 23 '16 at 11:13
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The term being limited is

$$\left (e^{\log{n}/n}-1 \right )^{1/n} = \left (\frac{\log{n}}{n} + \cdots\right )^{1/n}$$

So consider

$$\lim_{n \to \infty}\frac1n \log{\left (\frac{\log{n}}{n} \right )} = \lim_{n \to \infty} \frac{n}{\log{n}} \frac{1-\log{n}}{n^2} = 0$$

by L'Hopital. This is the log of the limit. The limit we seek is therefore $1$.

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Using $ \displaystyle 1+\frac1n \lt \sqrt[n]{n} \lt 1+\sqrt{\frac1n} $ for $n \ge 3$ and thus $ \displaystyle \frac{1}{ 1+\sqrt{\frac1n}} \lt \sqrt[n]{\frac1n} \lt \frac{1}{ 1+{\frac1n}}$

gives $ \displaystyle \frac1n \lt \sqrt[n]{n}-1 \lt \sqrt{\frac1n} $

so $ \displaystyle \frac{1}{ 1+\sqrt{\frac1n}} \lt \sqrt[n]{{\frac1n}} \lt \sqrt[n]{\sqrt[n]{n}-1} \lt \sqrt[n]{\sqrt{\frac1n}} \lt \sqrt{\frac{1}{ 1+{\frac1n}}}$

with the left- and right-hand expressions each converging towards $1 $ as $n$ increases

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  • $\begingroup$ I can prove that $\sqrt[n]{n} - 1 < \sqrt{\frac{2}{n}}$. But can you show me a proof that $\sqrt[n]{n} - 1 < \sqrt{\frac{1}{n}}$ too? $\endgroup$ – marmistrz Jan 23 '16 at 11:10
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    $\begingroup$ @marmistrz One approach is to look at $\left(1+\frac{1}{\sqrt{n}}\right)^n = 1 + \frac{n}{\sqrt{n}}+\frac{n(n-1)}{2 (\sqrt{n})^2}+ \frac{n}{\sqrt{n}}+\frac{n(n-1)(n-2)}{6 (\sqrt{n})^3}+\cdots \ge\frac{1}{3 \sqrt{n}} +\frac{1}{2}+\frac{\sqrt{n}}{2} +\frac{n}{2} + \frac{{{n}^{3/2}}}{6} $, with $\frac{n}{2} + \frac{{{n}^{3/2}}}{6} \ge n$ when $n\ge 9$ and the smaller values can easily be checked individually. Then take the $n$th root $\endgroup$ – Henry Jan 23 '16 at 13:53
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Idea: extend to $\Bbb R$, take $\log$: $$\lim_{x\to+\infty}\frac{\log(x^{1/x}-1)}x$$ and use L'Hôpital.

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Only Taylor series, only hardcore)

1) $n^{\frac{1}{n}} = e^{\frac{\log n}{n}} \sim 1 + \frac{\log n}{n}$, so the expression under the first root is $\frac{\log n }{n}$

2)$\big(\frac{\log n}{n} \big)^\frac{1}{n} = e^{\frac{\log \frac{\log n}{n}}{n}} = e^{\frac{\log \log n}{n}} \cdot e^{-\frac{\log n }{n}} \sim (1+\frac{\log \log n}{n})(1-\frac{\log n }{n}) \to_n 1$

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We can manipulate limits as long as we are sure that the part we are jumping over is bounded. For example:

$$\lim\limits_{x \to \infty} f(x)^{g(x)} = \lim\limits_{x \to \infty} f(x)^{\lim\limits_{x \to \infty}g(x)}$$

as long as $f(x)$ is bounded.

First we attack $\lim\limits_{n \to \infty} n^{\frac{1}{n}}$. We make a substitution $x=\frac{1}{n}$

$$\lim\limits_{n \to \infty} n^{\frac{1}{n}}=\lim\limits_{x \to 0} (\frac{1}{x})^x=\lim\limits_{x \to 0} \frac{1}{x^x}$$

Now this one is elementary $\lim\limits_{x \to 0} x^x = 1$ so we have $\lim\limits_{n \to \infty} n^{\frac{1}{n}}=1$

We continue

$$\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}-1}=\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}(1-\frac{1}{\sqrt[n]{n}})}=\lim\limits_{n \to \infty}n^{\frac{1}{n^2}} (1-\frac{1}{n^{\frac{1}{n}}})^{\frac{1}{n}}$$

Since $n^{\frac{1}{n^2}}=(n^{\frac{1}{n}})^{\frac{1}{n}}$ we have as well $\lim\limits_{n \to \infty} n^{\frac{1}{n^2}}=1$

That leaves

$$\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}-1}=\lim\limits_{n \to \infty} (1-(\frac{1}{n})^{\frac{1}{n}})^{\frac{1}{n}}$$

The same substitution and we have

$$\lim\limits_{x \to 0} (1-x^x)^{x}$$

$x^x$ has a minimum at $\frac{1}{e}$ and $1 > 1-x^x > 0$ for $0 < x < 1$ meaning it is bounded and $1-x^x$ is not reaching 0 for any $x$ in this region so we can examine

$$\lim\limits_{x \to 0} (1-x^x)^{x}=\lim\limits_{x \to 0} x^{x^2}(\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}$$

This is because we can write $x^{x^2}=(x^x)^x$

Obviously both limits are equal $\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (1-x^x)^{x}$.

Notice that

$$\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (x^{-x}-1)^{x}=\lim\limits_{x \to 0^{-}} (x^{x}-1)^{-x}=\lim\limits_{x \to 0^{-}} \frac{1}{(x^{x}-1)^{x}}$$

which is to say that the right limit is equal to reciprocal left (if it exists). This means that the unique limit, if it exists, is 1 since left and right limits are then the same.

But, $(1-x^x)^x$ is continuous at 0, $x^x$ is continuous and $a^x$ is continuous, meaning the limit is indeed 1.

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$$\lim _{n\to \infty \:}\left(\left(n^{\frac{1}{n}}-1\right)^{\frac{1}{n}}\right) = \lim _{n\to \infty \:}\left(e^{\frac{1}{n}\ln \left(n^{\frac{1}{n}}-1\right)}\right)$$ Apply the Limit Chain Rule $$\mathrm{if}\:\lim _{u\:\to \:b}\:f\left(u\right)=L,\:\mathrm{and}\:\lim _{x\:\to \:a}g\left(x\right)=b,\:\mathrm{and}\:f\left(x\right)\:\mathrm{is\:continuous\:at}\:x=b$$ $$\mathrm{Then:}\:\lim _{x\:\to \:a}\:f\left(g\left(x\right)\right)=L$$ $$g\left(n\right)=\frac{1}{n}\ln \left(n^{\frac{1}{n}}-1\right),\:f\left(u\right)=e^u$$ So $$\lim _{n\to \infty \:}\left(\frac{1}{n}\ln \left(n^{\frac{1}{n}}-1\right)\right)=\lim _{n\to \infty \:}\left(\frac{\ln \left(n^{\frac{1}{n}}-1\right)}{n}\right)$$ $\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{\frac{1-\ln \left(n\right)}{n^2-n^{2-\frac{1}{n}}}}{1}\right)=\lim _{n\to \infty \:}\left(\frac{-n^{\frac{1}{n}-2}\left(\ln \left(n\right)-1\right)}{n^{\frac{1}{n}}-1}\right)$ $\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{n^{\frac{1}{n}-4}\left(\left(\ln \left(n\right)-1\right)^2+n\left(2\ln \left(n\right)-3\right)\right)}{n^{\frac{1}{n}-2}\left(-\left(\ln \left(n\right)-1\right)\right)}\right)= \lim _{n\to \infty \:}\left(\frac{3n-\ln ^2\left(n\right)-2n\ln \left(n\right)+2\ln \left(n\right)-1}{n^2\left(\ln \left(n\right)-1\right)}\right)$ $\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{\frac{2-2\ln \left(n\right)}{n}-2\ln \left(n\right)+1}{n\left(2\ln \left(n\right)-1\right)}\right)=\lim _{n\to \infty \:}\left(\frac{n-2n\ln \left(n\right)-2\ln \left(n\right)+2}{n^2\left(2\ln \left(n\right)-1\right)}\right)$ $\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{-\frac{2}{n}-2\ln \left(n\right)-1}{4n\ln \left(n\right)}\right)=\lim _{n\to \infty \:}\left(\frac{-n-2n\ln \left(n\right)-2}{4n^2\ln \left(n\right)}\right)$ $\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{-2\ln \left(n\right)-3}{4n\left(2\ln \left(n\right)+1\right)}\right)$ $\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{-\frac{2}{n}}{8\ln \left(n\right)+12}\right)$

Then $$=\lim _{n\to \infty \:}\left(\frac{-1}{2n\left(2\ln \left(n\right)+3\right)}\right)$$

Now $$\lim _{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)}\mathrm{,\:where\:}\lim _{x\to a}g\left(x\right)\ne 0$$ $$\frac{\lim _{n\to \infty \:}\left(-1\right)}{\lim _{n\to \infty \:}\left(2n\left(2\ln \left(n\right)+3\right)\right)}=\frac{-1}{\infty \:}=0$$

So $$\lim _{u\to \:0}\left(e^u\right)=1$$

Finally $$\lim _{n\to \infty \:}\left(\left(n^{\frac{1}{n}}-1\right)^{\frac{1}{n}}\right)=1$$

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