Let $z$ be complex with $z=1$, $H(n)=\sum_{k=1}^n \frac{1}{k}$ be the $n$th harmonic number, denote $$S=\sum_{n=1}^N z^{H(n)}$$ then what can we say about $S$? Does there exist a good upper bound on this sum?

1$\begingroup$ Perhaps compare to $\sum_{n=1}^N z^{\log n} = \sum_{n=1}^N n^{\log z} < \zeta(\log z)$ $\endgroup$– Gerald EdgarAug 26 '17 at 1:20

1$\begingroup$ @GeraldEdgar: I don't think the inequality follows, or perhaps more to the point, what does it even mean? $\endgroup$– Christian RemlingAug 26 '17 at 1:37

$\begingroup$ Inequality requires $\log z > 1$ of course. So this is not the case $z = 1$. $\endgroup$– Gerald EdgarAug 26 '17 at 1:43

$\begingroup$ But $z$ is complex. $\endgroup$– Robert IsraelAug 26 '17 at 1:43

1$\begingroup$ Trivially, $S \le N$, with equality at $z=1$. $\endgroup$– Robert IsraelAug 26 '17 at 1:53
We write $z = e^{2\pi is}$ for $s \in \mathbf{R}$. Let $\gamma$ be the Euler constant so that $H(n) = \log n + \gamma + o(1)$. We claim that $$S = \frac{Ne^{2\pi is(\log N + \gamma)}}{2\pi i s + 1} + o(N).$$ Put $S_N = N^{1}\sum_{n=1}^N e^{2\pi i sH(n)}$ and $S'_N = N^{1}\sum_{n=1}^N e^{2\pi i s\log n}$.
Suppose we have managed to show that $S'_N = \frac{N^{2\pi is}}{2\pi i s + 1} + o(1)$. Then since $$e^{2\pi is \gamma}S_N'S_N \le N^{1}\sum_{n=1}^N\lefte^{2\pi is (\log n + \gamma)}(1  e^{2\pi i s(H(n)\log n  \gamma)})\right $$
Clearly, the term $1  e^{2\pi i s(H(n)\log n  \gamma)}$ is arbitrarily small when $n\to \infty$. Thus $$\lim_{N\to \infty}e^{2\pi is \gamma}S_N'S_N = 0,$$ and the claim will follow. It remains then to show that $$S'_N = \frac{N^{2\pi is}}{2\pi i s + 1} + o(1). $$ We shall calculate the limit $$ \lim_{N\to\infty}\frac{S'_N}{N^{2\pi i s}} = \lim_{N\to\infty}N^{1}\sum_{n=1}^N \left(\frac{n}{N}\right)^{2\pi i s} $$ For any $\epsilon > 0$, we choose a continuous function $\varphi_{\epsilon}(x)$ on $[0, 1]$ such that $0\le \varphi_{\epsilon} \le 1$, $\varphi_{\epsilon}(0) = 0$ and $\varphi_{\epsilon}\big_{[\epsilon, 1]} = 1$. Then we have the cut off limit $$ \lim_{\epsilon \searrow 0}\lim_{N\to \infty} N^{1}\sum_{n=1}^N \left(\frac{n}{N}\right)^{2\pi i s}\varphi_{\epsilon}\left(\frac nN\right) = \lim_{\epsilon \searrow 0}\int^1_0x^{2\pi is}\varphi_{\epsilon}(x)dx $$ By dominant convergence, $$\lim_{\epsilon \searrow 0}\int^1_0x^{2\pi is}\varphi_{\epsilon}(x)dx = \int^1_0x^{2\pi is} dx = \frac{1}{2\pi is + 1}.$$ Now consider the difference between the orginal limit and the cut off limit. The value $1  \varphi_{\epsilon}\left(\frac nN\right)$ is $0$ when $n/N \ge \epsilon$ and is $\le 1$ in general, so $$ \lim_{\epsilon \searrow 0}\lim_{N\to \infty} N^{1}\sum_{n=1}^N \left\left(\frac{n}{N}\right)^{2\pi i s}\left(1  \varphi_{\epsilon}\left(\frac nN\right) \right)\right \le \lim_{\epsilon \searrow 0}\lim_{N\to \infty} N^{1}\sum_{\substack{n=1 \\ n/N < \epsilon}}^N 1\le \lim_{\epsilon \searrow 0} \epsilon = 0, $$ Therefore, the two limits agree, giving us $$ \lim_{N\to\infty}N^{1}\sum_{n=1}^N \left(\frac{n}{N}\right)^{2\pi i s} = \lim_{\epsilon \searrow 0}\lim_{N\to \infty} N^{1}\sum_{n=1}^N \left(\frac{n}{N}\right)^{2\pi i s}\varphi_{\epsilon}\left(\frac nN\right) = \frac{1}{2\pi is + 1}. $$ We have now proven $\lim_{N\to\infty}S'_N/N^{2\pi i s} = (2\pi is + 1)^{1} $, whence $$S'_N = \frac{N^{2\pi i s}}{2\pi is + 1} + o(1)$$ as desired.

1$\begingroup$ I don't understand the end of your proof, I'm sorry. First, I don't get the convergence to the integral since your function cannot be continuously defined at 0. Second, you only prove that $S'_N=\frac{N^{2\pi is}}{2\pi is+1}+o(N)$. Could you provide more details ? $\endgroup$– M. DusAug 26 '17 at 15:26

$\begingroup$ The limit $\lim S'_N / N^{2\pi is} = (2\pi is + 1)^{1}$ really means $S'_N = N^{2\pi is}/(2\pi is + 1) + o(1)$, since $N^{2\pi is} = 1$. $\endgroup$ Aug 26 '17 at 15:32

$\begingroup$ I've added a cut off function $\varphi_{\epsilon}$ in order to apply the Riemann sum. I hope it works now. $\endgroup$ Aug 26 '17 at 16:36

$\begingroup$ @M.Dus to your first question, the function $x^{2\pi is}$ on $[0,1]$ is Lebesgue integrable, and its Lebesgue integral is equal to its improper Riemann integral. $\endgroup$ Aug 26 '17 at 17:35

$\begingroup$ Okay thank you and sorry for my thoughtless question. Of course is is $o(1)$. $\endgroup$– M. DusAug 27 '17 at 10:27