Question
I am trying to find a simple argument students not trained to university level maths that the series
$$\zeta(s) = \sum \frac{1}{n^s}$$
diverges for $\sigma=0$, where $s=\sigma+it$.
Here "simple" means avoiding Dirichlet series theory taught at university.
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My attempt
If we look at the terms we see that $$\left|\frac{1}{n^{s}}\right|=\left|\frac{1}{n^{\sigma}}\frac{1}{n^{it}}\right|=\left|\frac{1}{n^{\sigma}}\right|$$
Writing $n^{it}$ as $e^{it\ln(n)}$ makes clear it has a magnitude of 1.
When $\sigma<0$ the magnitude of the terms grow larger than 1, making it easy to conclude the series diverges for $s<0$.
However, for $s=0$, this isn't so easy. Each term is
$$\frac{1}{n^{\sigma}} \frac{1}{n^{it}} = e^{-it\ln(n)}$$
Although the magnitude of such terms is still 1, the rotation might mean they do or don't cancel.
One case of cancelling is
$$ \sum_n e^{-i \pi n} = 1 -1 +1 -1 + \ldots $$
which oscillates, definitely doesn't converge, and many would say doesn't diverge.
Q1: Can we say that because $t\ln(n)$ grows slower than $\pi n$, the cancelling will be insufficient to prevent divergence?
I then think of another case of cancelling
$$ \sum_n e^{-i\pi \frac{n}{2}} = 1 + i -1 -i +1 +i -1 -i + \ldots $$
Q2: Does the argument change to say that because $t\ln(n)$ grows slower than $\pi n/R$, for any real $R$, the cancelling will still be insufficient to prevent divergence?