4
$\begingroup$

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?

  • 1
    Your inequality $\frac1{1-n}\ge\ln(n)\ge1+n$ is false.2012-11-14

4 Answers 4

1

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 $-\infty$.

2

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.

1

$\log(x) \leq x - 1$, 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 $-\infty$.

1

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.

  • 0
    @PeterTamaroff: first of all, it's Jean Le Rond d’Alembert, not D Alambert. Then in the above messages I pointed out that I considered these limits as trivial. You meet them in the first pages in any calculus books, and that's why I just used the results without any further proof. Of course, this doesn't mean that it's a bad thing to post their proofs one billion (or more) times.2012-11-16