Let the Riemann Zeta function be expressed as: $$\zeta(x) = \sum_{n=1}^{\infty} \frac{1}{n^x}$$ I want to show, it is (1) continuous using that Weierstrass M-test, (2) but not uniformly continuous for $x \geq 1$. My proof goes like the following:
(1) Fix $\delta > 0$ and observe that the series $\sum_{n=1}^{\infty} \frac{1}{n^x}$ converges uniformly for $x \geq \delta + 1$ because $\sum_{n=1}^{\infty} \frac{1}{n^x} \leq \sum_{n=1}^{\infty} \frac{1}{n^{\delta +1}} < \infty$ and using Weierstrass M-test gives us uniform continuity. Then $\zeta(x)$ is continuous in $[\delta + 1, \infty)$ and we are done.
(2) I want to show that $\zeta(x)$ is unbounded for $x \geq 1$, ...
I am wondering if proof for (1) is correct and the answer for (2). Thanks!
Your proof of continuity on $(1,\infty)$ is correct. [ Note that you have written $x \geq 1$ instead of $x>1$.
If the function is uniformly continuous then it would be bounded on $(1,T]$ for any $T \in (0,\infty)$. However $\zeta (x) \geq \sum\limits_{n=1}^N \frac 1 {n^{x}} $ for each $N$. If $M$ is an upper bound for the function choose $N$ such that $\sum\limits_{n=1}^N \frac 1 n >M+1$ and then choose $x$ close enough to $1$ to get a contradiction.