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Determine all real values of $p$ for which the following series converges: $\displaystyle \sum_{n=1}^{\infty} \left(\sin\left(\dfrac{1}{n}\right)\right)^p$.

I am sort of confused how to solve this. What methods are there to find values of convergence and to prove that a series converges? I know that we could use the integral test and ratio test, but I don't think the integral test would work here since $\sin\left( \dfrac{1}{x}\right)$ isn't a decreasing function.

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    $\begingroup$ $\sin(1/n)$ behaves very much like $1/n$ when $n$ is large. And $\sum\frac1{n^p}$ converges exactly when $p>1$. Use the "limit comparison test". $\endgroup$ – Gregory Grant Dec 7 '15 at 18:09
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Hint for some values of p.

For large n, sin(1/n) is approximately 1/n. This simply leads to considreation of: $$\lim_{n\to\infty} n^p\sin^p(1/n)=1$$. So for p>1, the given series converges.

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  • $\begingroup$ It should be noted though that $\sin(\dfrac{1}{n}) > 0$ for $n>1$. $\endgroup$ – user19405892 Dec 7 '15 at 20:52

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