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Convergence of the following series as $\alpha\in\mathbb{R}$ $$\sum_{n=1}^{\infty}(-n)^{\lfloor\alpha\rfloor}\frac{\sin^3\left(\frac{1}{n}\right)}{\left(\log(n)\right)^{\alpha-1}}$$

As $n \to +\infty$ we have that $\frac{\sin^3\left(\frac{1}{n}\right)}{\left(\log(n)\right)^{\alpha-1}} = \mathcal{O}\left(\frac{1}{n^3\log^{\alpha-1}(n)}\right)$

How to handle $(-n)^{\lfloor\alpha\rfloor}$ with that result?

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  • $\begingroup$ i think you should take two cases, for $\lfloor\alpha\rfloor$ to be odd or even $\endgroup$ – Subhajit Halder Sep 6 '18 at 12:49
  • $\begingroup$ @F.inc Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE $\endgroup$ – gimusi Oct 23 '18 at 21:13
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Yes the key point is that

$$\frac{\sin^3\left(\frac{1}{n}\right)}{\left(\log(n)\right)^{\alpha-1}} \sim \frac{1}{n^3\log^{\alpha-1}n}$$

then we need to distinguish the cases, for example for $0\le \alpha <1$ we have $\lfloor\alpha \rfloor=0$ and then

$$(-n)^{\lfloor\alpha\rfloor}\frac{\sin^3\left(\frac{1}{n}\right)}{\left(\log(n)\right)^{\alpha-1}} \sim \frac{\log n}{n^3}$$

which converges by limit comparison test with $\sum \frac1{n^2}$.

And you can proceed similarly for the others values.

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