Limit:$ \lim\limits_{n\rightarrow\infty}\left ( n\bigl(1-\sqrt[n]{\ln(n)} \bigr) \right )$ 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?
 A: 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. 
A: $\log(x) \leq x - 1$, 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 $-\infty$.
A: 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.
A: 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 $-\infty$.
