# On the sum $\sum_{n=1}^N z^{H(n)}$

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?

• Perhaps compare to $\sum_{n=1}^N z^{\log n} = \sum_{n=1}^N n^{\log z} < \zeta(-\log z)$ Aug 26 '17 at 1:20
• @GeraldEdgar: I don't think the inequality follows, or perhaps more to the point, what does it even mean? Aug 26 '17 at 1:37
• Inequality requires $-\log z > 1$ of course. So this is not the case $|z| = 1$. Aug 26 '17 at 1:43
• But $z$ is complex. Aug 26 '17 at 1:43
• Trivially, $|S| \le N$, with equality at $z=1$. Aug 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 $$|e^{2\pi is \gamma}S_N'-S_N| \le N^{-1}\sum_{n=1}^N\left|e^{2\pi is (\log n + \gamma)}(1 - e^{2\pi i s(H(n)-\log n - \gamma)})\right|$$
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.
• 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 ? Aug 26 '17 at 15:26
• 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 $|N^{2\pi is}| = 1$. Aug 26 '17 at 15:32
• I've added a cut off function $\varphi_{\epsilon}$ in order to apply the Riemann sum. I hope it works now. Aug 26 '17 at 16:36
• @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. Aug 26 '17 at 17:35
• Okay thank you and sorry for my thoughtless question. Of course is is $o(1)$. Aug 27 '17 at 10:27