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Compute $$\lim_{n\to\infty}\frac{1}{n}(1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+ \sqrt[n]{n}-n)$$

$$\lim_{x\to\infty}\frac{x^{\ln x}}{{(\ln x)}^x}$$

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For second, top is $e^{(log x)^2}$, bottom is $e^{x\log\log x}$. Bottom wins big time. – André Nicolas Dec 27 '12 at 17:00
@AndréNicolas: thanks for your hint / way! The 2nd one seems rather troublesome and couldn't find anything nice/fast. – user 1618033 Dec 27 '12 at 17:05
You mean the first one, $(1/n)(\sum\sqrt[k]{n}-n)$? – André Nicolas Dec 27 '12 at 17:13
@AndréNicolas: No. I mean exactly what I wrote in my question. – user 1618033 Dec 27 '12 at 17:25
One can find more than one solution to the first question, for example by approximating by integral. It is just that there is a lot of slack, so much cruder estimate will do. – André Nicolas Dec 27 '12 at 17:57
up vote 1 down vote accepted

Calculation with brute force: $$\lim_{x\to\infty}\frac{x^{\ln x}}{{(\ln x)}^x}=\lim_{x\to\infty}\frac{e^{\ln^2 x}}{{e^{\ln(\ln x) x}}}=\lim_{x\to\infty}e^{\ln^2 x-x\ln(\ln x)} $$ Then, $$\lim_{x\to\infty}\ln^2 x-x\ln(\ln x)=\lim_{x\to\infty}\frac{\frac{\ln^2 x}x-\ln(\ln x)}{\frac1x}=\frac{0-\infty}{0^+}=-\infty $$ Thus, $$\lim_{x\to\infty}\frac{x^{\ln x}}{{(\ln x)}^x}=\lim_{x\to\infty}e^{\ln^2 x-x\ln(\ln x)}=0 $$

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@Chris'ssister Well then there is a mitake somewhere. – Nameless Dec 27 '12 at 17:26
@DonAntonio Yeah I noticed. Thanks – Nameless Dec 27 '12 at 17:28

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