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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.

Supriyo Banerjee
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  • See if you can exploit propositions 2.1 and 2.2 from here – rtybase Nov 25 '18 at 10:07
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    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. – xbh Nov 25 '18 at 13:19

1 Answers1

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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 $y \to \infty$.

user
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  • 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... – Supriyo Banerjee Nov 26 '18 at 03:08
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    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. – user Nov 26 '18 at 06:28