The series $\sum_{n=1}^\infty \frac{\sin(n\pi/3)}{\sqrt{n+1}}$ converges/diverges? 
Does the series $\sum_{n=1}^\infty \dfrac{\sin(\frac{n\pi}{3})}{\sqrt{n+1}}$ converge, converge absolutely or diverge?

I am again lost with a series.  With Dirichlet's test I came to the conclusion that it converges. But I don't know how to show that it doesn't converge absolutely. Any ideas?
 A: Hint: $$n\gt 1 \Rightarrow n > \sqrt n  \Rightarrow  \frac{1}{\sqrt n} >\frac1n$$
Now use the harmonic series.
Edit: I don't really know what you intend with considering $\sin^2(\cdots)$. It's much simpler than that. The idea is that there's subsequences where the term is a constant divided by $\sqrt n$ (the same constant for each term). For example take the series (which is "contained" in your series) $$\sum_{k=0}^{\infty}\frac{\sin\frac{(6k+1)\pi}{3}}{\sqrt{6k+2}} =\sum_{k=0}^{\infty}\frac{\sin{(2k\pi+\frac{\pi}{3}})}{\sqrt{6k+2}} = \sin(\pi/3)\sum_{k=0}^\infty \frac{1}{\sqrt{6k+2}}$$
which diverges by the first hint. Since the original series in absolute value is non-negative and greater than or equal to the above series, it too is divergent.
A: We have that
$$\sin\frac{n\pi}3=\begin{cases}\;\;\;\frac{\sqrt3}2&,\;\;n=1,2\pmod 6\\{}\\\;\;\;\;\;0&,\;\;n=0,3\pmod 6\\{}\\-\frac{\sqrt3}2&,\;\;n=4,5\pmod 6\end{cases}\implies\left|\sin\frac{n\pi}3\right|=0\,,\;\frac{\sqrt3}2$$
From here that
$$s_{3k}=\sum_{n=1}^{3k}\left|\frac{\sin{\frac{n\pi}3}}{\sqrt{n+1}}\right|=\frac{\sqrt3}2\left(\frac1{\sqrt1}+\frac1{\sqrt2}+\frac1{\sqrt4}+\frac1{\sqrt5}+\ldots+\frac1{\sqrt{3k-2}}+\frac1{\sqrt{3k-1}}\right)\ge$$
$$\ge\frac{\sqrt3}2\cdot\frac{2k}{\sqrt{3k-1}}\xrightarrow[k\to\infty]{}\infty$$
Thus the sequence of partial sums of the absolute values of the series isn't bounded above and thus the series diverges.
