Find for which $\alpha$ and $x$ the series converges $$ \sum_{n=1}^{\infty}=\frac{(\sin x)^{3n}}{(n+1)^\alpha}\hspace{2 ex}\alpha>0,\hspace{1 ex}x\in\mathbb{R}$$
My work
By the ratio test I have
$$\lim_{n→∞}\left|\frac{(\sin x)^{3n+3}}{(\sin x)^{3n}}\frac{(n+1)^\alpha}{(n+2)^\alpha }\right|=\left|(\sin x)^3\right|\left(\frac{n+1}{n+2}\right)^\alpha=|\sin x|^3$$
So the series converges if $|\sin x|^3<1\Leftrightarrow x\notin\left(\pi/2+2k\pi, k\in\mathbb{Z}\right)\cup\left(3\pi/2+2k\pi, k\in\mathbb{Z}\right)$
If $x\in\left(3\pi/2+2k\pi, k\in\mathbb{Z}\right)$ the series is
$$ \sum_{n=1}^{\infty}=\frac{(-1)^{3n}}{(n+1)^\alpha}$$
It follows by the alternating series test, the series converges
Am I right?
If $x\in\left(\pi/2+2k\pi, k\in\mathbb{Z}\right)$ the series converges?
Unless $|\sin(x)|=1$, the series converges (absolutely) by the ratio test for any values of $\alpha$. This means that it converges absolutely for any values of $x$ that are not in the form $2\pi k \pm \pi/2$.
If $x$ is in the form $2\pi k + \pi/2$, then $\sin(x)=1$ and we are left with a typical p-series, which we know to converge for all $\alpha > 1$.
If $x$ is in the form $2\pi k -\pi/2$, then $\sin(x)=-1$ and $(-1)^{3n}=(-1)^n$, so we have an alternating series. This converges since its terms alternate in sign, approach zero, and decrease in absolute value for all $\alpha > 0$.
In summary, the series converges if:
What you’ve written so far looks correct; you’ve just left out the case of $\sin(x)=1$.