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I have to show that for any positive rational number $r$, the sequence $\left\{\frac{\log n}{n^r}:n\ge1\right\}$ is bounded.

Is this correct argument: $\lim\limits_{n\to\infty}\frac{\log n}{n^r}=\lim\limits_{n\to\infty}\frac{1/n}{rn^{r-1}}=0$, i.e., convergent, hence bounded.

I am also asked to prove that $$\sum_{n\ge10}\frac{(\log n)^2(\log\log n)}{n^2}$$ converges using above info or otherwise.

I can't think f how to proceed.

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    $\begingroup$ I'm not sure if they wanted you to use (at least directly) L'Hopital's rule for the first problem. Note that $r$ being a positive rational is not used. Since it asks for the weak result of boundedness, I would suspect it wants you to use the fact that $\log n<n$ for all $n\in\mathbb N$. As for the series, can you replace the numerator with powers of $n$ in such a way that the resulting series still converges? $\endgroup$ – Simply Beautiful Art May 9 '19 at 14:02
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The first one is correct.

For the second one here is a hint (of course you can use the first exercise after choosing a small $r>0$). Since $\log x<x,\forall x>0$, if you have $x^ε$ with $ε>0$ instead of $x$, you get the inequality $\log x<\frac{1}{ε}x^ε,\ \forall ε>0.$ So, it is easy to check that $\log \log x<\log x<\frac{1}{ε}x^{\epsilon}, \forall ε>0$ too. Now, can you use these inequalities to compare your series to a convergent one?

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