What is the difference between the following $2$ sets? What is the difference between the following two sets?


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*$\{s\in\mathbb C:\Re(s)\ge1+\delta\},\quad\delta>0$

*$\{s\in\mathbb C:\Re(s)>1\}$
I read that $\displaystyle\sum\limits_{n\in\mathbb N}\frac{1}{n^s}$  is absolutely and uniformly convergent for the first one and normally convergent for the second one, as normal convergence implies uniform and absolute convergence the second set should be smaller, but is there an example where normal convergence fails for the first set?
 A: The difference is that in the first case some $\delta >0$ is fixed, and then a set $S_{\delta}= \{s\in\mathbb C:\Re(s)\ge 1+\delta\}$ is considered. This set is for every $\delta$ a strict subset of $S= \{s\in\mathbb C:\Re(s)>1\}$. For example, $S_{\delta}$ does not contain $1+\delta/2 $ or $1+(\delta/9) + 45i$ and so on.
Of course the union of $S_{\delta}$ over all $\delta$ is $S$; but this is something else. 
If you know the terms: note that the  sets $S_{\delta}$ is closed while the set $S$ is open. 
On your other question: It is not correct that the function is normally convergent for the second set, assuming that normal convergence has the meaning as on the linked page and the functions considered are $f_n(s) = n^s$. 
What is still true is that for every $s \in S$ the series is absolutely convergent. Perhaps there is some confusion about the meaning of normal convergence. It is also true that one has local normal convergence (normal convergence on a neighborhood for each point in the domain) and also compact normal convergence (normal convergence on every compact set in the domain). 
Indeed, the reference you provided now (Freitag and Busam, Comlex Analysis) uses the terminology "normal convergence" for what is "local normal convergence." And  normal convergence in this local sense, just does not imply uniform convergence in general. 
Moreover, one has in fact normal convergence for the first sets $S_{\delta}$. Possibly there was some mix-up with the conditions; 
but one does not even have uniform convergence for the second set.  
