Let $p>0$. I must find the values of $p$ for which the following series converges: $$\sum_{n=1}^{\infty}\frac{\sin n}{n^p}$$ I have already successfully proven the following estimate by induction: $$\sin 1+ \cdots+ \sin n=\frac{\sin\left(\frac{n+1}{2}\right)\sin\left(\frac{n}{2}\right)}{\sin\left(\frac12\right)}$$ But this means that the sequence $\sin 1 +\cdots + \sin n$ is bounded for all $n$. Then, the original series converges if $(a_n)_{n=1}^\infty=\frac{1}{n^p}$ is a decreasing sequence of positive numbers such that $\lim_{n\to\infty}=0$. Then, $(a_n)_{n=1}^\infty$ converges for all $p>0$, as this is the the only scenario which fulfills both conditions.
I have used the following theorem:
Let $\sum_{n=1}^\infty b_na_n$ be an infinite sequence such that:
- The sequence $B_n=b_1+\cdots+b_n$ is bounded.
- $(a_n)_{n=1}^\infty$ is a decreasing sequence of positive numbers such that $\lim_{n\to\infty}a_n=0$
Then, $\sum_{n=1}^\infty b_na_n$ is converging.
Now, when would $\sum_{n=1}^{\infty}\frac{\sin n}{n^p}$ converge absolutely? We would then have to find the values of $p$ for which the following converges: $$\sum_{n=1}^{\infty}\frac{\lvert \sin n\rvert}{n^p}$$ Would I still be able to use my original estimate for $\sin n$? If so, the answer would be the same as above, but I am not certain how to determine this.
Edit: While graphing variations of this, it seems that the above series converges for all $p>1$. Is there a formal way I can show that this is true?
Best Answer
As you found out, Dirichlet test takes care of convergence, whatever $p$.
Regarding absolute convergence, comparison test implies $\sum_{n\geq1}\frac{\sin n}{n^p}$ is absolutely convergent whenever $p>1$.
Concerning, $p=1$, the series diverges absolutely (see this How to prove that $ \sum_{n \in \mathbb{N} } | \frac{\sin( n)}{n} | $ diverges?)
This solves the rest of the problem: if $p<1$, $\left|\frac{\sin(n)}{n^p}\right|\geq | \frac{\sin( n)}{n} |$ and comparison test yields divergence.