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I am working on a problem which asks to prove whether the series $$\sum_{n=1}^\infty \sin(n^{-\alpha})$$ converges given $\alpha > 0$.

I am guessing that it depends on whether $\alpha > 1$ since $\sin(x)$ behaves like $x$ as $x \to 0$. Toward this end, I can use L'Hopital's rule to show that $$\lim_{n \to \infty} \frac{\sin(n^{-\alpha})}{n^{-\alpha}} = 1$$ which shows that in fact $\sin(n^{-\alpha}) \approx n^{-\alpha}$ for $n$ large.

My question is whether this is sufficient to prove the conjecture that it converges for $\alpha > 1$ (and doesn't converge otherwise), since the rate of convergence of $\sin(n^{-\alpha})$ to $n^{-\alpha}$ might also matter.

Thanks

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    Nope, you are good. – 2017-02-05
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    An alternative would be to use:$$\sin(n^{-\alpha}) – 2017-02-05
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    @Simply: It is important to note that for $n$ large enough, $\sin(n^{-\alpha})>0$. This is what makes that argument suffice. – 2017-02-05
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    I think the tool you're looking for is the [Limit Comparison Test](https://en.wikipedia.org/wiki/Limit_comparison_test). By the way, you don't need l'Hôpital's rule to evaluate that limit—just the facts that $\lim_{n\to\infty} n^{-\alpha} = 0$ and $\lim_{x\to0} \frac{\sin x}x = 1$. – 2017-02-05

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Note the for $0\le x\le \pi/2$, the sine function satisfies the inequlities

$$\frac{2x}{\pi}\le \sin(x)\le x \tag1$$

Substituting $x=n^{-\alpha}$ into $(1)$ gives

$$\frac{2}{\pi n^\alpha}\le \sin(1/n^\alpha)\le \frac{1}{n^\alpha}$$

for $\alpha >0$.

Hence, from the comparison test, the series converges for $\alpha >1$ and diverges for $0<\alpha \le 1$.