Prove the series $\sum n^{-1-it}$ is diverge for all real $t$.

Question

Prove that the series $\sum_{n=1}^\infty n^{-1-it}$ diverges for all real $t$.

I have shown in the previous exercise that this series is bounded for nonzero $t$, and when $t=0$, it is famous that the harmonic sum is divergent. I know various techniques to prove a series convergent, but I don't know how to prove a series is divergent.

I have used Abel's Summation Formula, Euler Summation Formula, prove by contradiction, but these cannot get me to the result. Any suggestion?

Answer 1

You can use a general theorem about Dirichlet series,

If $F(s) = \sum_{n = 1}^{\infty} \frac{a_n}{n^s}$ converges for $s_0 = \sigma_0 + i t_0$, then $$\lim_{\sigma \downarrow \sigma_0}\; (\sigma - \sigma_0)F(\sigma + it) = 0$$ for all $t \in \mathbb{R}$.

In particular $F$ cannot have a pole on the line $\sigma = \sigma_0$.

Since $\zeta(s)$ has a pole at $1$ it follows that $\sum_{n = 1}^{\infty} \frac{1}{n^{1+it}}$ diverges for all $t\in \mathbb{R}$.

This argument shows the divergence without any computation, but it doesn't show in which way the series diverges for $s = 1+it$, and of course the theorem must be known for the argument to be legitimate.

We can determine the way in which the series diverges by using one of Abel's sum formula, Euler's sum formula, or the Euler–Maclaurin sum formula (and probably in other ways too). Taking Abel's formula, for $t \neq 0$ we find \begin{align} \sum_{n = 1}^{N} 1\cdot\frac{1}{n^{1+it}} &= \frac{N}{N^{1+it}} + (1+it) \int_{1}^{N} \frac{\lfloor u\rfloor}{u^{2+it}}\,du \\ &= \frac{N}{N^{1+it}} + (1+it) \int_{1}^{N} \frac{u - \lbrace u\rbrace}{u^{2+it}}\,du \\ &= N^{-it} + (1+it)\int_{1}^{N} \frac{du}{u^{1+it}} - (1+it)\int_{1}^{N} \frac{\lbrace u\rbrace}{u^{2+it}}\,du \\ &= -\frac{N^{-it}}{it} + \frac{1+it}{it} - \int_{1}^{\infty} \frac{\lbrace u\rbrace}{u^{2+it}}\,du + \int_{N}^{\infty} \frac{\lbrace u\rbrace}{u^{2+it}}\,du\,. \end{align} In this, $$\frac{1+it}{it} - \int_{1}^{\infty} \frac{\lbrace u\rbrace}{u^{2+it}}\,du$$ is a constant (namely $\zeta(1+it)$), and $$\Biggl\lvert \int_{N}^{\infty} \frac{\lbrace u\rbrace}{u^{2+it}}\,du \Biggr\rvert \leqslant \int_{N}^{\infty} \frac{du}{u^2} = \frac{1}{N}$$ tends to $0$. The term $-\frac{N^{-it}}{it}$ moves (clockwise) around a circle with radius $1/t$.

Thus altogether the partial sums spiral towards the limiting circle $\lvert z - \zeta(1+it)\rvert = 1/t$ and every point of this circle is an accumulation points of partial sums.