Testing for the convergence of the following series Determine if the following series is convergent:
$$\sum_{k=1}^\infty \frac{\sin^3 k}{\sqrt k}.$$
At first when I am solving the problem, I thought of the limit comparison test, if I divide the term by $\frac{k^3}{\sqrt k}$. Then I will get a limit of 1, and since the summation of the term I divide is divergent, I conclude that it is divergent. 
However, I think that there is something wrong because if I do the same thing on $\frac{\sin^3 k}{k^2}$ by dividing it by $\frac{k^3}{k^2}$, I also get a limit of 1 but summation of k clearly diverges. This is not right because sum $\frac{\sin^3 k}{k^2}$ clearly converges. What is wrong here?

The series is convergent: we may express $\sin^3j$ as a linear combination of $\sin(3j)$ and $\sin j$, so we are reduced to prove that the series 
$$\sum_{j=1}^\infty\frac{\sin(3j)}{\sqrt{j}}\mbox{ and }\sum_{j=1}^\infty\frac{\sin j}{\sqrt j}$$
are convergent. 
To this aim, we use a summation by part plus boundedness of the sequences 
$\left(\sum_{j=1}^n\sin(3j)\right)_{n\geqslant 1}$ and $\left(\sum_{j=1}^n\sin j\right)_{n\geqslant 1}$.
