Converge or diverge? $\alpha 0, \alpha \in \mathbb R$
$$\sum_{n=1}^{\infty} sin (n^{-\alpha})$$

Question

Converge or diverge? $\alpha >0, \alpha \in \mathbb R$
$$\sum_{n=1}^{\infty} sin (n^{-\alpha})$$

rational · Answer

recall that for $x\gt 0$ we have $\sin(x)\lt x$
from this we get
$$\sin(n^{-\alpha}) \lt n^{-\alpha}$$

since $\sum\limits_{n=1}^{\infty}n^{-\alpha}$ converges for $\alpha\gt 1$, the series $\sum\limits_{n=1}^{\infty} \sin (n^{-\alpha})$ also converges for $\alpha \gt 1$.

we still need to see the behavior for $0\lt\alpha\le 1$

loser66 · Answer

we have as t --> 0, sint / t --->1, can we use it?

rational · Answer

for testing the convergence in interval (0, 1] ?

loser66 · Answer

yes. If we use -1 < sin t < 1 where t = 1/ n ^ alpha

rational · Answer

Ahh are you allowed to use limit comparison test ?

rational · Answer

$a_n=\sin(\frac{1}{n^{\alpha}})$

$b_n = \frac{1}{n^{\alpha}}$

$$\lim\limits_{n\to\infty}\dfrac{a_n}{b_n} =\lim\limits_{n\to\infty}\dfrac{\sin(\frac{1}{n^{\alpha}})}{\frac{1}{n^{\alpha}}}=1 $$

therefore the series in question will have same behavior as p-series. that is, the series covnerges for $\alpha\gt 1$ and diverges for $\alpha\le 1$

loser66 · Answer

Yes,

loser66 · Answer

Wow!! how easy it is to you. Thank you so much. I do appreciate.

rational · Answer

sinx/x=1 limit made it easy :) thanks to you too!

loser66 · Answer

I need help understanding the proof. Please theorem 1.2 http://www.ux1.eiu.edu/~kparwani/integrals.pdf