Find the limit of $n \ (\sqrt[n]{n} - 1) $

Question

Find the limit of $n \ (\sqrt[n]{n} - 1) $.

Wolfram says it is $+\infty$.

Obviously $n \rightarrow +\infty$ and $\sqrt[n]{n} - 1\rightarrow 0$.

Any hint?

Intelligenti pauca · Accepted Answer · 2015-11-13T15:28:18.980

13

Hint $$ n \ (\sqrt[n]{n} - 1)={e^{(\ln n)/n}-1\over (\ln n)/n}\ln n. $$

edited Nov 13 '15 at 15:28

answered Nov 13 '15 at 13:46

Intelligenti pauca

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I don't get this transformation. – tomtom Nov 13 '15 at 14:13
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$\sqrt[n]{n}=n^{1/n}=\exp(\ln(n^{1/n}))=\exp({1\over n}\ln n)$ – Intelligenti pauca Nov 13 '15 at 14:22
but how did you get $ln {\frac{n}{n}}$ ? – tomtom Nov 13 '15 at 14:47
The parentheses are missing, it's $\frac{\ln n}{n}$. – Clement C. Nov 13 '15 at 14:57
Ok. I see. And $\frac{e^{(\ln n)/n}-1}{ (\ln n)/n}\ln n$ diverges to what? – tomtom Nov 13 '15 at 15:04
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Look at the two factors: $\ln n$ goes to infinity; and $\frac{e^h-1}{h}=\frac{e^h-e^0}{h-0}\to \exp^\prime(0) = e^0 = 1$ (note that $h=\frac{\ln n}{n}\to 0$). – Clement C. Nov 13 '15 at 15:07

score 2 · Answer 2 · answered Nov 13 '15 at 14:22

With Taylor approximations, using the facts that (i) $e^x = 1+x +o(x)$ when $x\to 0$, (ii) $\sqrt[n]{n}=e^{\frac{\ln n}{n}}$, and (iii) ${\frac{\ln n}{n}}\xrightarrow[n\to\infty]{}0$:

(See below for a full, detailed derivation; stop here if you just wanted a hint.)

$$\begin{align}n\left(\sqrt[n]{n}-1\right) &= n\left(e^{\frac{\ln n}{n}}-1\right)= n\left(1+\frac{\ln n}{n} + o\left(\frac{\ln n}{n}\right)-1\right) \\&= n\left(\frac{\ln n}{n} + o\left(\frac{\ln n}{n}\right)\right)= \ln n + o(\ln n)\end{align}$$and since $\ln n \xrightarrow[n\to\infty]{} \infty$ you can conclude by comparison.

I'll come back here after I cover Tayler approximation in calculus class :) — tomtom, Nov 13 '15 at 14:54

Arashium · Answer 3 · 2015-11-20T09:06:20.850

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$$\lim_{n\to\infty}\frac{(n^\frac1n-1)}{\frac1n}$$

Using Hopital

$$=\lim_{n\to\infty}\frac{n^{\frac1n}\frac1{n^2}\times (-\ln n +1)}{\frac{-1}{n^2}}$$ $$=\lim_{n\to\infty}n^{\frac1n}(\ln n -1)$$

edited Nov 20 '15 at 09:06

answered Nov 13 '15 at 13:58

Arashium

2,541

What exactly are your "=" signs referring to? This should be the limits that are equated, yet here the parsing would suggest it's the actual quantities. – Clement C. Nov 13 '15 at 14:28
@ClementC. I mean the second phrase and the third one are equal to the first phrase. – Arashium Nov 15 '15 at 21:03
No -- their limits are equal but they are not themselves equal. – Clement C. Nov 15 '15 at 21:05
@ClementC. Nice point. Thanks – Arashium Nov 20 '15 at 09:06

score 0 · Answer 4 · answered Nov 13 '15 at 15:12

$$n(\sqrt[n] n-1)= \frac {\sqrt[n] n-1}{\frac{1}{n}}$$ Then applying L'Hôpital's rule, one has (after simplification in numerator and denominator of the factor $\frac{1}{n^2}$ which I can't arrive to write in Tex Commands) $$\frac{-n^{1/n}(\ln n-1)}{-1}$$ and this expression clearly tends to $+\infty$ when $n\to\infty$

Find the limit of $n \ (\sqrt[n]{n} - 1) $

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