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?
Best Answer
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.