This question is from Serge Lang's textbook, in a chapter that comes before the ratio and integral tests are introduced, so those can't be used. I've already proved that $\sum\frac{\log n}{n^3}$ converges and have an inkling that this result may be useful, but I can't figure out how.
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1How did you prove that $\sum \frac{\log n}{n^3}$ converges? – Daniel Fischer Aug 15 '15 at 14:34
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$\sum\frac{log n}{n^3}$ < $\sum\frac{n}{n^3}$, which converges. – bard Aug 15 '15 at 14:38
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Okay. So if you could show that $(\log n)^2 \leqslant n$, you'd be done. – Daniel Fischer Aug 15 '15 at 14:40
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Hint: Prove $(\log n)^2 < n$ for large enough $n$ and compare your series to $\sum\frac{1}{n^2}$.
ChocolateBar
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$$ \forall \varepsilon>0, \,\, \sum_{n\ge 1}\frac{1}{n^3}<\sum_{n\ge 1}\frac{(\ln n)^2}{n^3} \le \sum_{n\ge 1}\frac{1}{n^{3-\varepsilon}}+O(1). $$
Paolo Leonetti
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