For what values of $a$ does $\sum_{n=1}^\infty \left( 1+\frac12 + \dotsb + \frac1n \right) \frac{\sin (na)}{n}$ converge?

Your reasoning looks sound. When in doubt, you can always do summation by parts to check; for instance, in this case, with $H_n = 1+\cdots + 1/n$ and $s_n = \sum_{k = 1}^n \sin{(an)}$, you would write \begin{align*} \sum_{n=1}^N {H_n \sin{(na)}\over n} & = {H_Ns_N\over N}+\sum_{n=1}^{N-1}\left({H_n\over n} -{H_{n+1}\over n+1}\right)s_n\\ & = {H_Ns_N\over N}+\sum_{n=1}^{N-1}{(n+1)H_n-nH_{n+1}\over n(n+1)} s_n. \end{align*} Since $(n+1)H_n - nH_{n+1} = H_n-1$ and $s_n$ is bounded, the above partial sum can be compared to a sum with terms $H_n/n^2\approx \log{n}/n^2$ which converges.