3
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ Perhaps compare to $\sum_{n=1}^N z^{\log n} = \sum_{n=1}^N n^{\log z} < \zeta(-\log z)$ $\endgroup$ Aug 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? $\endgroup$ Aug 26 '17 at 1:37
  • $\begingroup$ Inequality requires $-\log z > 1$ of course. So this is not the case $|z| = 1$. $\endgroup$ Aug 26 '17 at 1:43
  • $\begingroup$ But $z$ is complex. $\endgroup$ Aug 26 '17 at 1:43
  • 1
    $\begingroup$ Trivially, $|S| \le N$, with equality at $z=1$. $\endgroup$ Aug 26 '17 at 1:53
2
$\begingroup$

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.

$\endgroup$
6
  • 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. Dus
    Aug 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 $|N^{2\pi is}| = 1$. $\endgroup$
    – Wille Liou
    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. $\endgroup$
    – Wille Liou
    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. $\endgroup$
    – Wille Liou
    Aug 26 '17 at 17:35
  • $\begingroup$ Okay thank you and sorry for my thoughtless question. Of course is is $o(1)$. $\endgroup$
    – M. Dus
    Aug 27 '17 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.