Dirichlet series - distance between $\sigma_u$ and $\sigma_c$ In a book available here we see the following 
In summary the above says: if a Dirichlet series is convergent at $s_0$, it is uniformly convergent in a compact sector right of it. 
Also if the Dirichlet series is convergent at $s_0$ it is convergent at every $s: \{\Re(s) > \Re(s_0)\}$.
The definition of abscissa of uniform convergence being: 
Now putting all this together I am confused about why $\sigma_u \ne \sigma_c$. The Dirichlet series is convergent at all $s$ in the half plane defined by $\Re(s) > \sigma_c$ and all the $z$ in the line say $\Re(z) = \sigma_c+\epsilon$ as $s_0$ we should be able to get compact right plane $\Re(s) \ge \sigma_c+2\epsilon$ where the series is uniformly convergent. Setting $\delta =2\epsilon$ arbitrarily small, $\sigma_u$ should be arbitrarily close to $\sigma_c$ if not equal. 
But clearly, that does not seem to be the case, reading further in the book. 
Where am I making a mistake? Is it that you cannot merge all the locally uniformly convergent regions? If yes, what are the additional conditions -if any are possible- under which such regions can be merged? 
 A: Note first that if $f_n \to f$ uniformly on some set $D$ (doesn't need to be compact, open etc) and for all $n$ (high enough), $f_n$ is bounded by some $M_n$, then $f_n, f$ are uniformly bounded (uniform covergence is equivalent to uniform Cauchy, apply this with $\epsilon =1$, so there is $N_0, |f_n(x)| \le |f_{N_0}(x)|+1 \le M_{N_0}+1=M$ for all $x \in D, n \ge N_0$ and then $|f(x)| \le M+1$ by the same reasoning). 
Let's assume we have regular Dirichlet series from now on as the results are clearer this way, but similar results hold for generalized ones. Let's call $\sigma_c, \sigma_a, \sigma_u, \sigma_b$ the abscissas of convergence, absolute convergence, uniform convergence and bounded convergence (this last means that $f$ is bounded in $\Re s > \sigma_b + \epsilon$). 
The above result shows that $\sigma_u \ge \sigma_b$ since the partial sums of Dirichlet functions are (individually) bounded by the (finite) sum of the absolute value of the coefficients when $\Re s >0$ and by a translation we can always assume that $\sigma_c \ge 0$, so we can discuss everything in the right half-plane as again the results are clearer this way. 
Conversely assuming $\sigma_c \ge 0$ and letting $U_N=\sup_t{|\sum_{n \le N} {a_n n^{-it}}}| \le \sum_{n\le N} {|a_n|}$, it is obvious summing by parts that if $U_N \le N^r$ for all $N$ high enough, the partial series converge uniformly on $\Re s > r +\epsilon$ hence on $\sigma_b + \epsilon$ so $\sigma_u \le \sigma_b$, hence they are both equal to $\limsup \frac{\log U_N}{\log N}$. Since $\sigma_a =\limsup \frac{\log A_N^*}{\log N}$, where as usual $A_N^*=\sum_{n\le N}{|a_n|}$ and since $\sum_{n \le N}{|a_n|^2}=\lim_{T \to \infty}\frac{1}{2T}\int_{-T}^T|\sum_{n\le N}{a_n n^{-it}|^2}dt\le U_N^2$, Cauchy-Schwarz shows that $A_N^* \le \sqrt N U_N$, hence $\sigma_a \le \sigma_u + \frac{1}{2}$, while we also know that $\sigma_a \le \sigma_c+1$
In particular for the Dirichlet eta (the alternate RZ) we have $\sigma_u = \sigma_b= \sigma_a=1, \sigma_c=0$ since RZ is unbounded at infinity on the line $\sigma =1$ as for $|t| \pm \infty$, we have $|\zeta(1+it)| \ge C\log \log t$), hence the eta function is unbounded there, so this is a simple example of where $\sigma_u=\sigma_c+1$ which is the maximum possible, while for any series where $\sigma_c=\sigma_a$ we have the equality with $\sigma_u$. It is non-trivial to construct series where we have the other extreme, $\sigma_a=\sigma_u +\frac{1}{2}$ but such have been known since the 1930's.
So if $\sigma_a > \sigma_c+\frac{1}{2}$ (so always in the maximal case $\sigma_a=\sigma_c+1$) we will never have $\sigma_c=\sigma_u$, while if $\sigma_c=\sigma_a$ we obviously have $\sigma_u$ equal to them. In between it depends on the series but as noted examples can be constructed with pretty much everything allowed by these general inequalities.
(edited per question asked in comment)
Uniform convergence in a sector centered on the abscissa of convergence and with opening $\alpha < \pi$ doesn't mean uniform convergence on the whole half-plane even if you go a little more to the right - just draw a picture to see that even if your sector angle is very close to $\pi$ you will fail to cover the (two) sectors that form the angle $(\pi -\alpha)/2$ "up" and "down" with the vertical line; those will actually intersect any half-plane to the right, so that's why finding half-planes of uniform convergence is hard unless they come from half-planes of absolute convergence (which implies uniform convergence), since absolute convergence is just a coefficient matter (so no issue with the angles if you want) as the Dirichlet series with positive coefficients obviously satisfies $|f(\sigma+it)| \le f(\sigma)$; you have uniform convergence on compacts, but right half-planes are not so even if they contain the line bounding them to the left, as they are unbounded vertically and to the right; local uniform approximations (ie on compacts) and global uniform approximations (on the whole open set) are very, very different issues  as one can see ths easily on the open unit disc- any analytic function there is locally uniformly approximated by polynomials (take their taylor series), but the ones globally approximated are the ones that extend continuosly to the closed unit disc so in particular they cannot be unbounded, hence something like $\frac{1}{1-z}$ cannot be globally approximated there
