Show that the series is divergent

Question

I have the following series: $$ \sum\limits_{n=1}^{\infty} \frac{\sin(n)}{3n^p-\sin(n)} $$ and I want to show that this series diverges for $0<p \leqslant \frac 1 2$. I tried to write $$ \frac{\sin(n)}{3n^p - \sin(n)} = \frac{\sin(n)}{3n^p} \left(1-\frac{\sin(n)}{3n^p}\right)^{-1} = \frac{\sin(n)}{3n^p}\left(1+\frac{\sin(n)}{3n^p}+\cdots \right) \\ = \frac{\sin(n)}{3n^p} \left(1+\frac{\sin(n)}{3n^p}+\cdots+\frac{\sin^k(n)}{(3n^p)^{k}} \left(1-\frac{\sin(n)}{3n^p}\right)^{-1}\right) \\ = \frac{\sin(n)}{3n^p} + \frac{\sin^2(n)}{(3n^p)^{2}}+\cdots+\frac{\sin^k(n)}{(3n^p)^k} + \frac{\sin^{k+1}(n)}{(3n^p)^k \bigl( 3n^p-\sin(n) \bigr)} $$ and then choose for any fixed $p$ such $k$ that $pk > 1$ so that the series related to last term converge. The series related to first term converges for any $p>0$. So there is a problem to show that the series, related to group of terms from first to $k$-th diverges. For each of these terms I can show this by expanding $\sin^m(n)$ as a sum of sines and cosines in first power and constants. But how can I show that the resulting sum will correspond to an also divergent series? Maybe there is some better way to show that my series diverges?

Answer 1

Summation by parts gives the convergence of $\sum_n\frac{\sin n}{n^p}$, so we have to deal with the convergence of $\sum_n a_n$, where $a_n:=\sin n\left(\frac 1{3n^p-\sin n}-\frac 1{3n^p}\right)=\frac{\sin^2n}{3n^p(3n^p-\sin n)}$.

We have a series with non negative terms, and $a_n\sim\frac{\sin^2n}{9n^{2p}}$.

Rewriting $\sin^2n$ as $1-\cos(2n)$ (up to a multiplicative constant), noticing that $\sum_n\frac{\cos(2n)}{n}$ is convergent and $\sum_n\frac 1{n^{2p}}$ is divergent, we conclude the divergence of the initial series.