Uniform convergence of $\sum_{k=1}^\infty \frac{1}{k^{1+x}}$ Problem:
Show that $\sum_{k=1}^\infty \frac{1}{k^{1+x}}$ converges uniformly on $[a,\infty)$ for any $a > 0$, but does not converge uniformly on $(0,\infty)$.

what I have done:
Let $f_n = \sum_{k=1}^n \frac{1}{k^{1+x}}$ for $n \in \mathbb{N}$.
By the Cauchy criterion for uniform convergence, $\sum_{k=1}^\infty \frac{1}{k^{1+x}}$ converges uniformly on $[a,\infty)$ for $a > 0$ iff for all $\epsilon > 0$ and $x \in [a,\infty)$ there exists $n_0 \in \mathbb{N}$ such that $|f_m(x) - f_n(x)| = |\sum_{k=1}^n \frac{1}{k^{1+x}} - \sum_{k=1}^n \frac{1}{k^{1+x}}| = \sum_{k=m+1}^n \frac{1}{k^{1+x}}  < \epsilon$ for all $n,m \ge n_0$ (with $m < n$).
For every $x \in [a,\infty)$, $x \le a \Rightarrow k^{1+x} \ge k^{1+a} \Rightarrow \frac{1}{k^{1+x}} \le \frac{1}{k^{1+a}} $ for $k \in \mathbb{N}$. Then $f_n(x) \le f_n(a)$ for all $n \in \mathbb{N}$.
Then $\sum_{k=m+1}^n \frac{1}{k^{1+x}} \le \sum_{k=m+1}^n \frac{1}{k^{1+a}} < \sum_{k=m+1}^\infty \frac{1}{k^{1+a}}$.
So, given $\epsilon$ and $a$ we want to find $n_0$ such that $\sum_{k=n}^\infty \frac{1}{k^{1+a}} < \epsilon$ for all $n \ge n_0$.

To show that $\{f_n\}$ does not uniformly converge on $(0,\infty)$, it is sufficient to show that $\{f_n\}$ does not uniformly converge on $(0,1) \subset (0,\infty)$.
Fix $\epsilon > 0$. We must show that for all $n \in \mathbb{N}$ there exists $x \in (0,\infty)$ such that $|f_n(x) - f(x)| = \sum_{k=n+1}^\infty \frac{1}{k^{1+x}} \ge \epsilon$. 

To show non-uniform convergence (last part of your post), bound the remainder series from below by an integral $$\int_{n+1}^\infty  \frac{dy}{y^{1+x}}  $$
