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

To my thinking, $$f(n)=\frac{\left( 1+\frac12 +
 \dotsb + \frac1n \right)}{n}$$ will behave like $\frac{\log n}{n}$, which is not convergent, but certainly has terms going to zero (eventually monotonically). By Dirichlet's test, $$\sum_{n=1}^\infty f(n) \sin(an)$$ will converge provided that the partial sums of $\sum_{n=1}^\infty \sin(na)$ are bounded, which they always are. So it should converge for all $a\in \mathbb{R}$.
Would appreciate anyone pointing out mistakes in this reasoning. Thanks
 A: 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.
A: Some caution is required. The observation that "$A$

will behave like

$B$" is useful as a heuristic method to shape one's expectations, but it can be misleading. One always needs to check whether $A$ behaves similarly enough to $B$ that the conclusion about $B$ carries over.
In the situation of the question that is the case, but that requires more than just that
$$H_n = \sum_{k = 1}^{n} \frac{1}{k}$$
behaves like $\log n$. If in
$$\sum_{n = 1}^{\infty} H_n \frac{\sin (na)}{n} \tag{1}$$
we replace $H_n$ with $g(n)$ where $g(n)$ "behaves like" $\log n$ in the sense that $\lvert g(n) - \log n\rvert$ remains bounded, the conclusion that the series converges for all $a\in \mathbb{R}$ need not hold. For example with $g(n) = \log n + \sin n$ the series
$$\sum_{n = 1}^{\infty} g(n)\frac{\sin (na)}{n}$$
diverges for $a = \pm 1$.
But of course $H_n$ behaves "more like $\log n$" than $g(n)$ does in so far as $H_n - \log n \to \gamma$ while $g(n) - \log n = \sin n$ oscillates. However, this is not enough to carry over the conclusion either. If for $n \geqslant 2$ we take $h(n) = \log n + \gamma + \frac{\sin n}{\log n}$ then we have $h(n) - \log n \to \gamma$ too, but still
$$\sum_{n = 2}^{\infty} h(n)\frac{\sin (na)}{n}$$
diverges for $a = \pm 1$.
Of course actually $H_n$ is tied much closer to $\log n$ than $h(n)$ is, namely we have
$$H_n = \log n + \gamma + O\bigl(n^{-1}\bigr)\,,$$
and the $O\bigl(n^{-1}\bigr)$ term decays fast enough to make the series
$$\sum_{n = 1}^{\infty} \bigl(H_n - \log n - \gamma\bigr)\frac{\sin (na)}{n}$$
absolutely convergent, so that we can deduce the convergence of $(1)$ for all $a$ from the convergence of the two series
$$\sum_{n = 1}^{\infty} \frac{\sin (na)\log n}{n} \qquad\text{and}\qquad \sum_{n = 1}^{\infty} \frac{\sin (na)}{n}\,.$$

But we can also directly obtain the convergence of $(1)$ for all $a \in \mathbb{R}$ using Dirichlet's test:
\begin{align}
\frac{H_n}{n} - \frac{H_{n+1}}{n+1} &= \frac{H_n}{n} - \frac{H_n}{n+1} - \frac{1}{(n+1)^2} \\
&= \frac{H_n}{n(n+1)} - \frac{1}{(n+1)^2} \\
&\geqslant \frac{1}{n(n+1)} - \frac{1}{(n+1)^2} \\
&> 0
\end{align}
shows the monotonicity (right from the start), and $H_n/n \to 0$ is easy to show.
A: 
For what values of $\;a\in\mathbb R\;$ does $$\sum_{n=1}^\infty\left(1+\frac12+\ldots+\frac1n\right)\frac{\sin\!\big(na\big)}{n}$$ converge ?

First of all , we will prove that
if $\;a\ne2k\pi\;,\;\forall k\in\mathbb Z\;,\;$ then
$\displaystyle\sum_{n=1}^N\sin\!\big(na\big)=\frac12\bigg(\!1-\cos\!\big(Na\big)\!\bigg)\cot\left(\frac a2\right)+\frac12\sin\!\big(Na\big)\quad\color{blue}{(*)}$
for all $\;N\in\mathbb N\;.$
By using Prosthaphaeresis identities , we get that
$\sin\!\big(na\big)+\sin\big((n+1)a\big)=$
$\quad=2\sin\left(\dfrac{(n+1)a+na}2\right)\cos\left(\dfrac{(n+1)a-na}2\right)=$
$\quad=\!2\!\sin\left(\!\dfrac{(n\!+\!1)a\!+\!na}2\!\right)\!\sin\left(\!\dfrac{(n\!+\!1)a\!-\!na}2\!\right)\!\cot\left(\!\dfrac{(n\!+\!1)a\!-\!na}2\!\right)\!=$
$\quad=\bigg(\!\!\cos\!\big(\!na\!\big)-\cos\!\big((n+1)a\big)\!\!\bigg)\cot\left(\!\dfrac a2\!\right)\;,\;\;$ for any $\;n\in\mathbb N\,.$
$\displaystyle2\!\sum_{n=1}^N\sin\!\big(na\big)\!=\!\!\!\sum_{n=1}^{N-1}\!\!\bigg(\!\!\sin\!\big(\!na\!\big)\!+\!\sin\!\big((n\!+\!1)a\big)\!\!\bigg)\!+\!\sin a\!+\!\sin\!\big(\!Na\!\big)\!=$
$\displaystyle\quad=\!\!\sum_{n=1}^{N-1}\!\!\bigg(\!\!\cos\!\big(\!na\!\big)\!-\!\cos\!\big((n\!+\!1)a\big)\!\!\bigg)\cot\left(\!\dfrac a2\!\right)\!+\!\sin a\!+\!\sin\!\big(\!Na\!\big)\!=$
$\quad=\!\bigg(\!\!\cos a\!-\!\cos\!\big(\!Na\!\big)\!\!\bigg)\!\cot\left(\!\dfrac a2\!\right)\!+\!2\sin^2\!\!\left(\!\dfrac a2\!\right)\!\cot\left(\!\dfrac a2\!\right)\!+\!\sin\!\big(\!Na\!\big)\!=$
$\quad=\bigg(\!\!\cos a-\cos\!\big(\!Na\!\big)+2\sin^2\!\!\left(\!\dfrac a2\!\right)\!\!\bigg)\cot\left(\!\dfrac a2\!\right)+\sin\!\big(\!Na\!\big)\!=$
$\quad=\bigg(\!\!\cos a-\cos\!\big(\!Na\!\big)+1-\cos a\!\bigg)\cot\left(\!\dfrac a2\!\right)+\sin\!\big(\!Na\!\big)\!=$
$\quad=\bigg(\!1-\cos\!\big(\!Na\!\big)\!\bigg)\cot\left(\!\dfrac a2\!\right)+\sin\!\big(\!Na\!\big)\;,\;\;$ for any $\;N\!\in\mathbb N\,.$
Hence ,
$\displaystyle\sum_{n=1}^N\sin\!\big(na\big)=\frac12\bigg(\!1-\cos\!\big(Na\big)\!\bigg)\cot\left(\frac a2\right)+\frac12\sin\!\big(Na\big)$
for all $\;N\in\mathbb N\;.$
Consequently , from $\;(*)\;,\;$ it follows that
$\text{if }\;a\ne2k\pi\;,\;\forall k\in\mathbb Z\;,\;$ then
$\displaystyle\left|\sum_{n=1}^N\sin\!\big(na\big)\right|\leqslant\left|\cot\left(\frac a2\right)\right|+\frac12\;,\;\;\text{ for all }\;N\in\mathbb N\;.\quad\color{blue}{(**)}$
Let $\;a_n\!=\!\sin\!\big(na\big)\;,\;b_n\!=\!\dfrac1n\left(\!1\!+\!\dfrac12\!+\!\ldots\!+\!\dfrac1n\!\right)\;,\;\;$ for all $\;n\in\mathbb N\,,$
and let $\;M=\begin{cases}\;\left|\cot\left(\dfrac a2\right)\right|+\dfrac12\qquad\text{if }\;a\ne2k\pi\;,\;\forall k\in\mathbb Z\\\\\qquad\dfrac12\qquad\qquad\;\;\text{ if }\;\exists k\in\mathbb Z\;\big|\;a=2k\pi\qquad.\end{cases}$
It results that
$\text{if }\;a\ne2k\pi\;,\;\forall k\in\mathbb Z\;,\;$ then $\;\displaystyle\left|\sum_{n=1}^N a_n\right|\leqslant M\;,\;\text{for all }\,N\!\in\mathbb N\,,$
$\text{if }\;\exists k\in\mathbb Z\;\big|\;a=2k\pi\;,\;$ then $\;\displaystyle\left|\sum_{n=1}^N a_n\right|=0\;,\;\;\text{ for all }\;N\!\in\mathbb N\,.$
In any case ( for any $\;a\in\mathbb R\;$) ,$\;$ it results that
$\displaystyle\left|\sum_{n=1}^N a_n\right|\leqslant M\;,\;\;\text{ for all }\;N\in\mathbb N\;.\quad\color{blue}{(1)}$
Moreover ,
$b_n=\dfrac1n\left(1+\dfrac12+\ldots+\dfrac1n\right)=$
$\quad=\dfrac1{n+1}\left(1+\dfrac1n\right) \left(1+\dfrac12+\ldots+\dfrac1n\right)>$
$\quad>\dfrac1{n+1}\left(1+\dfrac12+\ldots+\dfrac1n+\dfrac1n\right)>$
$\quad>\dfrac1{n\!+\!1}\left(\!1\!+\!\dfrac12\!+\!\ldots\!+\!\dfrac1n\!+\!\dfrac1{n+1}\!\right)=b_{n+1}>0\;,\;\;\;\forall n\!\in\mathbb N\,.$
Hence ,
$b_n>b_{n+1}>0\;,\quad\text{ for all }\;n\in\mathbb N\;.\quad\color{blue}{(2)}$
Since $\;0<1+\dfrac12+\ldots+\dfrac1n\leqslant1+\ln n\;,\;\;\;\forall n\in\mathbb N\;,$
it follows that
$0<b_n\leqslant\dfrac{1+\ln n}n\;,\;\;$ for all $\;n\in\mathbb N\;.$
On the other hand , $\;\lim\limits_{n\to\infty}\dfrac{1+\ln n}n=0\;,\;$ hence , by applying the Squeeze Theorem , we get that
$\lim\limits_{n\to\infty}b_n=0\;.\quad\color{blue}{(3)}$
From $\;(1)\;,\;(2)\;$ and $\;(3)\;,\;$ by applying Dirichlet’s Test , it follows that the series
$\displaystyle\sum_{n=1}^{\infty}a_nb_n=\sum_{n=1}^\infty\left(1+\frac12+\ldots+\frac1n\right)\frac{\sin\!\big(na\big)}{n}$
converges for any $\;a\in\mathbb R\;.$
