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

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.

• Your series isn't positive so how come you tried those tests ? – DonAntonio Feb 29 '16 at 12:33

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