0
$\begingroup$

What is the limit of this sequence where , $U_n = \frac{(\log n )^p}{n}$ where $p \ge 0$.

I have done this problem when $p$ is an integer. Sorry but I am not too much familiar with writing questions in stack exchange.

$\endgroup$
  • $\begingroup$ See if you can exploit propositions 2.1 and 2.2 from here $\endgroup$ – rtybase Nov 25 '18 at 10:07
  • 1
    $\begingroup$ If you know the results for $p \in \mathbb N$, note that for general $p$, there is a $k\in \mathbb N$ s.t. $k \leqslant p<k+1$, and to get the limit, try the squeezing theorem. $\endgroup$ – xbh Nov 25 '18 at 13:19
0
$\begingroup$

We have that

$$\frac{(\log n )^p}{n}=e^{p\log(\log n)-\log n} \to 0$$

indeed

$$p\log(\log n)-\log n=\log n\left(p\frac{\log(\log n)}{\log n}-1\right)\to -\infty$$

since

$$\frac{\log(\log n)}{\log n} \to 0$$

which can be easily proved by $\frac{\log x}x \to 0$ as $x \to \infty$ by $x=\log y$ and $y \to \infty$.

$\endgroup$
  • $\begingroup$ Why the downvote, something wrong? $\endgroup$ – user Nov 25 '18 at 18:53
  • $\begingroup$ Can we write $lim$ $x tends to infinity$ $f(n)g(n)$ = $lim$ $x tends to infinity$ $f(n)$ × $lim$ $x tends to infinity$ $g(n)$. When one of the sequence diverges? I don't know... $\endgroup$ – Supriyo Banerjee Nov 26 '18 at 3:08
  • 1
    $\begingroup$ Note that here we are using something different that is $$\lim_{n\to \infty} f(n)=\lim_{n\to \infty} g(n)\cdot h(n)$$ and using the fact that $g(n)\to \infty$ and $h(n)\to -1$ and that’s leads to a not indeterminate form. $\endgroup$ – user Nov 26 '18 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.