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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

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    $\begingroup$ Looks good to me... $\endgroup$ – Igor Rivin Jan 2 '14 at 21:55
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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.

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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.

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