Apostol's IANT Section 11.6 is on "The half-plane of convergence of a Dirichlet Series".
In it he proves that if a Dirichlet series
- is bounded at $s_0$ then it is also bounded at $\sigma>\sigma_0$ (Lemma 2)
- converges at $s_0$ then it converges at $\sigma>\sigma_0$ (Theorem 11.8 using Lemma 2).
Images reproduced below to aid readers. Here we follow the traditional notation $s=\sigma+it$.
Question: My confusion is as follows.
- Lemma 2 is only about boundedness, not convergence.
- However Apostol extends Lemma 2 to show that the series $\left|\sum_{a<n\leq b}f(n)s^{-s} \right| \leq Ka^{\sigma_0-\sigma}$ converges, because the RHS tends to $0$ as $a\rightarrow\infty$. Here $K$ is independent of $a$.
My conclusion is then that if a series is bounded at $s_0$ then it is bounded at any $s$ with $\sigma>\sigma_0$ (lemma 2), and by Apostol's extended argument, it also converges at any $s$ with $\sigma>\sigma_0$. This seems wrong. What have I misunderstood?
To add to my confusion, $\sigma-\sigma_0$ can be as small as we want, which suggests a series converges if it is bounded even at $\sigma$.
As far as I can see, the first part of your argument is correct. The proof of Theorem $11.8$ uses only boundedness, not convergence. The second part, that boundedness at a point implies convergence at that point is incorrect.
Let $$F(s)=\sum_{n=1}^\infty f(n)n^{-s},$$ where $f(n)$ is defined, rather informally, as follows. We consider the partial sums when $s=1$.
$f(1) = 1$. So long as the partial sum so far is less than $2$, we define $f(n)=1$. So $f(2)=f(3)=f(4)=1$. Now $1+\frac12+\frac13+\frac14>2$, and we define $f(n)=-1$ until the partial sum is $<-2$, when we reverse course again, defining $f(n)=1$ again, until we exceed $2$, and so on.
Clearly, the partial sums are bounded, but the series does not converge. I think the point you are missing is that when $\sigma=\sigma_0$, we would have a division by $0$ in the bound in the lemma, so that the proof breaks down.