Let $p$, $q \in \mathbb{R}$ and see the series $$ \sum_{n=2}^{\infty} \frac{1}{n^p(\ln n)^q} $$ View with the comparison criterion that if $p> 1$ then the series is convergent for all $q$, and if $p < 1$, it is divergent for all $q$.

Can anyone help me getting started?

  • $\begingroup$ We have $\ln x\le\dfrac{x^\varepsilon-1}\varepsilon$ and $\ln x\le\dfrac{x^\varepsilon}{e\varepsilon}$ for every $\varepsilon>0$. (With equality iff $x=1$ and $x=e^{1/\varepsilon}$ respectively.) $\endgroup$ – Akiva Weinberger Apr 30 '15 at 13:48

Note that $$(\log n)^{q} = o(n^{p})$$ as $n \to \infty$ for all $p, q > 0.$


Hint: We have
$$\sum_{n=2}^{\infty} \frac{1}{n^{\alpha}} <\infty \Leftrightarrow \alpha > 1$$ and $$\ln(n) < n \forall ~ n \geq 1$$

  • 1
    $\begingroup$ Not enough to conclude, I guess... $\endgroup$ – Siminore Apr 30 '15 at 12:37

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