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I have a trouble with this example:

$$n(\sqrt[n]n-1)$$

I've been trying to do it this this way:

$$a_n = \frac{(\sqrt[n]n-1)(\sqrt[n]n^{n-1}+\sqrt[n]n^{n-2}+\dots+1)}{\sqrt[n]n^{n-1}+\sqrt[n]n^{n-2}+\dots+1}= \frac{n-1}{\sqrt[n]n^{n-1}+\sqrt[n]n^{n-2}+\dots+1}$$

And then I tried to factor out "n", but I realized that's not going to work here. Can you give me any tips? Thank you in advance.

2 Answers2

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We have that

$$n^{m}(\sqrt[n]n-1)=n^{m}(e^{\frac{\log n}n}-1)=n^{m}\frac{\log n}n\frac{e^{\frac{\log n}n}-1}{\frac{\log n}n}=\frac{\log n}{n^{1-m}}\cdot\frac{e^{\frac{\log n}n}-1}{\frac{\log n}n}\to 0\cdot 1=0$$

user
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  • Hello, does ~ means approximation here? Anyway, thank you for your help! –  Nov 15 '18 at 22:06
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    @iforgotmypass I've changed it by a derivation with standard limits in order to do not use asympthotics. – user Nov 15 '18 at 22:07
  • one more question, how can I prove that n^(m-1) * ln n --> 0? –  Nov 15 '18 at 23:02
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    One way is by l'Hopital from $\lim_{x\to \infty} \frac{\log x}{x^a}=\lim_{x\to \infty} \frac{\frac1x}{ax^{a-1}}$ – user Nov 15 '18 at 23:05
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    Otherwise by $x=e^y\to \infty$ we obtain $$\frac{\log x}{x^a}=\frac{y}{e^{ay}}$$ – user Nov 15 '18 at 23:06
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Consider the functions $$ f(x)=\frac{1-x^x}{x^m} $$ and $g(x)=f(x)/x^x$. Then the limit of $g$ at $0$ exists if and only if the limit of $f$ at $0$ exists and in this case they're equal, because $\lim_{x\to0}x^x=1$. (Write limits for $x\to0^+$, if you prefer or your conventions dictate it.)

Since $n^m(\sqrt[\scriptstyle n]{n}-1)=g(1/n)$, the limit you're looking for is the same as $$ \lim_{x\to0}f(x)=\lim_{x\to0}\frac{x^x(1+\log x)}{mx^{m-1}} $$ if it exists (the step is an application of l'Hôpital).

Now, $\lim_{x\to0}x^{1-m}=0$ (because $1-m>0$) and $$ \lim_{x\to0}x^{1-m}\log x=\frac{1}{1-m}\lim_{x\to0}x^{1-m}\log(x^{1-m})=0 $$ by the well known fact that $\lim_{x\to0}x\log x=0$.

egreg
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