Comparison test for trig series I have following series
$$\sum_{n=1}^\infty \frac{\sin(3n)}{n^4} $$
Then I use comparison test, compare them to
$$\sum_{n=1}^\infty \frac{1}{n^4} $$
And conclude they converge.
However, I have this task marked as mistake "You can do comparison test only on positive number series". So are there alternative ways to prove convergence?
I have tried symbolab but it also uses comparison test
 A: You can compare the series $\displaystyle\sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert$ with $\displaystyle\sum_{n=1}^\infty\frac1{n^4}$ and deduce from this that the series $\displaystyle\sum_{n=1}^\infty\frac{\sin(3 n)}{n^4}$ converges absolutely. Therefore, it converges.
A: We can't use direct comparison test directly since the $a_n$ term of the given series oscillates.
What we can use is the absolute convergence criterion that is
$$\sum |a_n|<\infty \implies \sum a_n<\infty$$
and in this case if we consider
$$\sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert$$
we can apply direct comparison test on that since $|a_n| \ge 0$ and we obtain that
$$\sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert \le \sum_{n=1}^\infty \frac{1}{n^4}$$
thus $\sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert$ converges and by absolute convergence criterion also the original series converges.
To summarize the steps for the proof are
$$ \sum_{n=1}^\infty \frac{1}{n^4}<\infty \stackrel{D.C.T.}\implies \sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert <\infty \stackrel{Abs.C.C.}\implies \sum_{n=1}^\infty \frac{\sin(3 n)}{n^4} <\infty$$
