Convergence of Dirichlet series Let $\alpha$ be a Dirichlet series such that $\alpha(s)$ converges for all real numbers $s>0$. Suppose that $\lim_{s\downarrow0}\alpha(s)$ exists. How to prove that $\alpha(0)$ converges?
I know this is true if and only if $\alpha(s)$ converges uniformly in $(0,1)$.
The abscissa of convergence is at most $0$. If $\sigma_c<0$, then $\alpha(0)$ converges. But if $\sigma_c=0$, then $0$ lies on the line of convergence so it is not sure whether $\alpha(0)$ converges.
 A: As is mentioned in an answer linked by Dap in a comment, the result you are trying to prove is false. For this, consider the series
$$\alpha(s)=\sum_{n=1}^\infty\frac{1}{n^{s+1+i}}=\zeta(s+1+i).$$
The series converges absolutely for $s>0$ (in fact, for all $s$ with $\operatorname{Re}(s)>0$), but it diverges at $s=0$. Here is an elementary proof.
Consider the series
$$\sum_{n=1}^\infty\int_n^{n+1}\frac{\mathrm dx}{x^{1+i}}=\lim_{n\rightarrow\infty}\int_1^n\frac{\mathrm dx}{x^{1+i}}.$$
For $n\leq x\leq n+1$ we have
$$\left|\frac{1}{x^{1+i}}-\frac{1}{n^{1+i}}\right|=\left|1+i\right|\left|\int_n^x\frac{\mathrm dt}{t^{2+i}}\right|\leq 2\int_n^x\frac{\mathrm dt}{t^2}\leq 2\int_n^{n+1}\frac{\mathrm dt}{n^2}=\frac{2}{n^2},$$
hence
$$\left|\int_n^{n+1}\frac{\mathrm dx}{x^{1+i}}-\frac{1}{n^{1+i}}\right|=\left|\int_n^{n+1}\left(\frac{1}{x^{1+i}}-\frac{1}{n^{1+i}}\right)\mathrm dx\right|\leq\int_n^{n+1}\left|\frac{1}{x^{1+i}}-\frac{1}{n^{1+i}}\right|\mathrm dx\leq\frac{2}{n^2}.$$
Since the series $\sum_{n=1}^\infty n^{-2}$ converges, this gives that $\sum_{n=1}^\infty n^{-(1+i)}$ converges iff the series I've defined above does. But we have
$$\int_1^n\frac{\mathrm dx}{x^{1+i}}=\frac{1}{-i}\left(\frac{1}{n^i}-1\right)=\frac{1}{-i}\left(e^{-i\log n}-1\right)$$
which does not converge as $n\rightarrow\infty$. Therefore the series defining $\alpha(0)$ does not converge.
