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 $|f_n(x) - f(x)| = \sum_{k=n+1}^\infty \frac{1}{k^{1+x}} \ge \epsilon$. 

Hint: As $x \to 0$, your series approaches the divergent series $\sum_n 1/n$ so you should be able to show the convergence is not uniform. If you can't show it directly easily, bound the series above and below by multiples of the Cauchy condensation criterion series for a sequence of positive decreasing terms which in your case is $\sum_m 2^{m}/(2^m)^{1+x}$, which will give you a geometric series whose partial sums you can easily evaluate and bound to disprove uniform convergence as $x \to 0$. It looks like you already have a proof for uniform convergence on $[a,\infty)$ for $a > 0$ if you just note that the remainder term is bounded by the value at $a$ for $a' \in [a,\infty]$, and the series is convergent for $a > 0$. 
