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I find to difficult to evaluate with $$\lim_{n\rightarrow\infty}\left ( n\left(1-\sqrt[n]{\ln(n)} \right) \right )$$ I tried to use the fact, that $$\frac{1}{1-n} \geqslant \ln(n)\geqslant 1+n$$ what gives $$\lim_{n\rightarrow\infty}\left ( n\left(1-\sqrt[n]{\ln(n)} \right) \right ) \geqslant \lim_{n\rightarrow\infty} n(1-\sqrt[n]{1+n}) =\lim_{n\rightarrow\infty}n *\lim_{n\rightarrow\infty}(1-\sqrt[n]{1+n})$$ $$(1-\sqrt[n]{1+n})\rightarrow -1\Rightarrow\lim_{n\rightarrow\infty}\left ( n\left(1-\sqrt[n]{\ln(n)} \right) \right )\rightarrow-\infty$$Is it correct? If not, what do I wrong?

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    $\begingroup$ Your inequality $$\frac1{1-n}\ge\ln(n)\ge1+n$$ is false. $\endgroup$ – Cameron Buie Nov 14 '12 at 20:27
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Since it's so hard let's solve it in one line

$$\lim_{n\rightarrow\infty}\left ( n\left(1-\sqrt[n]{\ln(n)} \right) \right )=-\lim_{n\rightarrow\infty}\left(\frac{\sqrt[n]{\ln(n)}-1}{\displaystyle\frac{1}{n}\ln(\ln (n)) }\cdot \ln(\ln (n))\right)=-(1\cdot \infty)=-\infty.$$

Chris.

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    $\begingroup$ This is a tad unclear. $\endgroup$ – Pedro Tamaroff Nov 16 '12 at 12:31
  • $\begingroup$ It seems you're using $$\lim_{x \to 0}\frac{\log(1+x)}{x}=1$$, for example, and that $$(\log n)^{1/n}\to 1$$, which would be good to leave explicit, so that every one may understand. $\endgroup$ – Pedro Tamaroff Nov 16 '12 at 13:50
  • $\begingroup$ @ Peter Tamaroff: since they are trivial limits I considered that everybody understand that. Anyway, I appreciate your message that makes my answer more clearer (if there is need for this). $\endgroup$ – user 1357113 Nov 16 '12 at 13:54
  • $\begingroup$ Ok. However, they are not "trivial" limits. In fact, I'm not sure about the second one. $\endgroup$ – Pedro Tamaroff Nov 16 '12 at 13:56
  • $\begingroup$ @PeterTamaroff: the second one? It's d'Alembert criterion that is evaluation just at a first sight. $\endgroup$ – user 1357113 Nov 16 '12 at 13:58
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Hint: We look at the behaviour of $$x\left(1-\sqrt[x]{\log x}\right)$$ for large $x$. Rewrite the expression as $$\frac{1-e^{\frac{\log\log x}{x}} }{\frac{1}{x}}.$$ Top and bottom both approach $0$ as $x\to\infty$, so the conditions for using L'Hospital's Rule hold. The rest is a calculation.

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$\log(x) \leq x - 1$, so $\log(x) = n \log(x^{1/n}) \leq n (x^{1/n} - 1)$ for all integral $n \geq 1$. Take $x = \log(n)$ to get $\log(\log(n)) \leq n(\log(n)^{1/n}-1)$ or $n(1-\log(n)^{1/n}) \leq -\log(\log(n))$. This shows that your limit is $-\infty$.

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Use Taylor!

$$n(1-\sqrt[n]{\log n}) = n (1-e^{\frac{\log\log n}{n}}) \approx n\left(1-\left(1+\frac{\log\log n}{n}\right)\right) = - \log\log n$$

which clearly tends to $-\infty$.

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