Suppose $\sum_{n=1}^\infty f(n)$ converges to a constant $A$, then prove that the Dirichlet series with coefficients $f(n)$ also converges if $\sigma>0$.
This can be proved easily if the series $\sum_{n=1}^\infty f(n)$ converges absolutely. However, I have no clue about how to prove this if the series only converges. I tried to fit this series to Cauchy's condition: for every $\varepsilon>0$ we claim that there exists a $p$ such that $$\left|\frac{f(n)}{n^s}+\cdots+\frac{f(n+p)}{(n+p)^{s}}\right|<\varepsilon$$ But this still requires the series of $f(n)$ to converge absolutely. So how can I approach this? This is Exercise 2 of Chapter 11 in Introduction to Number Theory by Apostol.
One proves that convergence of $\sum_{n=1}^\infty f(n)$ implies uniform convergence for $\sigma > 0, |\arg s| \le \alpha < \frac{\pi}{2}, s=\sigma+it$ while absolute convergence follows in general only for $\sigma > 1$.
Since for any $s, \Re s >0$ we can find an alpha (depending on $\Im s$ in general) as above, we get the required convergence
The proof is standard and uses that
$|n^{-s}-m^{-s}|=|\int_{\log m}^{\log n}se^{-sx}dx| \le |s||\int_{\log m}^{\log n}e^{-\sigma x}dx|=\frac{|s|}{\sigma}(m^{-\sigma}-n^{-\sigma})$ where $n \ge m \ge 1, s=\sigma+it, \sigma >0$
So as long as $|\frac{|s|}{\sigma}|=|\frac{1}{\cos \arg s}| \le c$ we can apply summation by parts to show that the Cauchy sums $\sum_{m}^{n}f(n)n^{-s} \to 0, n \ge m \to \infty$ and we actually have uniform convergence to zero as long as $c$ is fixed, or if you want $|\cos \arg s| \ge \delta >0$ fixed, which is precisely the angle condition $|\arg s| \le \alpha < \frac{\pi}{2}$