Does $\sum_{n=1}^\infty {\frac{ \sin(\frac{n\pi}{6})}{\sqrt {n^4 + 1}}}$ converge?

Question

Does the series $$\sum_{n=1}^\infty {\frac{ \sin(\frac{n\pi}{6})}{\sqrt {n^4 + 1}}}$$ converge?

I tried with comparison test, limit comparison test, ratio test and others but I cannot figure this out.

Answer 1

I think comparison test is useful. Since $|\sin x| \le 1$ and $\sqrt{n^4+1}\ge n^2$, $$ 0\le\left|\frac{\sin\frac{n\pi}{6}}{\sqrt{n^4+1}}\right| \le \frac{1}{n^2} $$ and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ converges. Therefore, given series converges absolutely.

Answer 2

$$\left|\frac{\sin\left(\frac{n\pi}{6}\right)}{\sqrt{n^4 +1}}\right|\leq\frac{1}{n^2}$$