How to prove the divergence of $\zeta(s)$?

Question

I was hoping someone could give me a proof of the divergence of the zeta series ($\sum_{n=1}^{\infty}\frac{1}{n^s}$) for $0<\Re(s)\leq1$. I understand how to prove the divergence if $s\in\mathbb{R}$, however I am not sure about how to approach this with complex values of $s$.

Answer 1

If $\sigma=\mathfrak{R}(s)\leq 0$, then by the Test for Divergence, we have divergence of the zeta series.

If $s=1$, then the zeta series is just the harmonic series. Thus, it diverges.

If $0<\sigma \leq 1$ and $s\neq 1$,

We apply partial summation, which is also integration by part for Stieltjes integrals, to the partial sum of the zeta series. Let $\{t\}=t-\lfloor t\rfloor$ be the fractional part of $t$. If $N\geq 2$, $0<\sigma\leq 1$ and $s=\sigma+it\neq 1$, then

$$ \begin{align} \sum_{n\leq N} \frac1{n^s} &= \int_{1-}^N \frac{d\lfloor t\rfloor}{t^s}\\ &=\frac{\lfloor t \rfloor}{t^s}\Bigg\vert_{1-}^N + s\int_{1-}^N \frac{\lfloor t \rfloor}{t^{s+1}} dt\\ &=\frac{t-\{ t\} }{t^s}\Bigg\vert_{1-}^N + s\int_{1-}^N \frac{t-\{t\}}{t^{s+1}} dt\\ &=N^{1-s}+O\left(\frac 1{N^{\sigma}}\right)+ \frac{st^{1-s}}{1-s}\Bigg\vert_{1-}^N -s\int_1^N \frac{\{t\}}{t^{s+1}}dt\\ &=N^{1-s}+O\left(\frac1{N^{\sigma}}\right)+\frac s{1-s} N^{1-s}+\frac s{s-1}-s\int_1^N \frac{\{t\}}{t^{s+1}} dt\\ &=N^{1-s}+\frac s{1-s} N^{1-s}+\frac s{s-1}-s\int_1^{\infty} \frac{\{t\}}{t^{s+1}} dt+O\left(\frac{1+|s|/\sigma}{N^{\sigma}}\right). \end{align} $$ The expression below is an analytic continuation of $\zeta(s)$ to $\sigma>0$. $$ \zeta(s)=\frac s{s-1}-s\int_1^{\infty} \frac{\{t\}}{t^{s+1}} dt. $$ Thus, we have for $0<\sigma\leq 1$, $$ \sum_{n\leq N}\frac1{n^s} = \frac {N^{1-s}}{1-s} + \zeta(s) + O\left( \frac{1+|s|/\sigma}{N^{\sigma}}\right). $$

For $0<\sigma<1$, we have $$ \left|\sum_{n\leq N}\frac1{n^s}\right|\geq \frac{N^{1-\sigma}}{|1-s|}-|\zeta(s)|+O\left( \frac{1+|s|/\sigma}{N^{\sigma}}\right)\rightarrow \infty \ \ \textrm{ as }\ N\rightarrow\infty. $$ . Thus, the zeta series diverges in this case, and the partial sums are unbounded.

With $\sigma =1$, $t\neq 0$, $s=1+it$, we have $$ \sum_{n\leq N} \frac 1{n^{1+it}} = \frac{N^{-it}}{-it} + \zeta(1+it) +O\left(\frac {1+|t|}N\right). $$

The $N$-th partial sum of the zeta series at $s=1+it$ with $t\neq 0$, shows an oscillating behavior due to $N^{-it}=\exp(-it \log N)$. The partial sums are bounded. Therefore, the zeta series diverges at $s=1+it$.