Riemann zeta function as a series

Example

For a complex number $s$ with real part $\sigma = \operatorname{Re} s$, consider the series $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}, \qquad n^{-s} = e^{-s \log n}.$$ It is the Dirichlet series of the constant arithmetic function $a(n) = 1$. Its terms satisfy $\abs{n^{-s}} = n^{-\sigma}$. By the p-series test, $\sum_n n^{-\sigma}$ converges exactly when $\sigma > 1$. So the series converges absolutely for every $s$ with $\operatorname{Re} s > 1$.

Fix $\sigma_0 > 1$; on the half-plane $\operatorname{Re} s \ge \sigma_0$ the bound $\abs{n^{-s}} \le n^{-\sigma_0}$ holds. So the Weierstrass M-test, in its form for complex-valued functions, shows that the series converges uniformly there. On the whole half-plane $\operatorname{Re} s > 1$ the convergence is not uniform. Indeed, for real $s \to 1^+$ the tail $\sum_{n > N} n^{-s}$ tends to $\sum_{n > N} n^{-1} = \infty$. So for every $N$ the tail is unbounded on the half-plane. At $s = 1$ the series is the harmonic series, which diverges.