[proofplan] Subtract the two Dirichlet series and prove that every coefficient of the difference is zero. If not, choose the first nonzero coefficient $c_m$ and multiply the identity by $m^s$. The remaining terms have factors $(m/n)^s$ with $n > m$, and absolute convergence at one fixed real point gives a summable majorant, so the tail tends to zero as $s \to \infty$. The limit then forces $c_m = 0$, contradicting the choice of $m$. [/proofplan] [step:Pass to the difference Dirichlet series] Define the complex sequence $(c_n)_{n \in \mathbb{N}}$ by \begin{align*} c_n := a_n - b_n \end{align*} for each $n \in \mathbb{N}$. For every $s \in \mathbb{C}$ with $\operatorname{Re}(s) > \sigma_0$, the series \begin{align*} \sum_{n=1}^{\infty} c_n n^{-s} \end{align*} converges absolutely, because \begin{align*} \sum_{n=1}^{\infty} |c_n n^{-s}| \leq \sum_{n=1}^{\infty} |a_n n^{-s}| + \sum_{n=1}^{\infty} |b_n n^{-s}|. \end{align*} For every real $s > \sigma_0$, absolute convergence permits termwise subtraction, and the assumed equality gives \begin{align*} \sum_{n=1}^{\infty} c_n n^{-s} = 0. \end{align*} [/step] [step:Choose the first nonzero coefficient and rescale the identity] Suppose, toward a contradiction, that not all coefficients $c_n$ vanish. Since $\mathbb{N}$ is well ordered, there exists a least index $m \in \mathbb{N}$ such that $c_m \neq 0$. Thus $c_n = 0$ for every $n < m$. For every real $s > \sigma_0$, the identity from the previous step becomes \begin{align*} 0 = \sum_{n=m}^{\infty} c_n n^{-s}. \end{align*} Multiplying by the positive real number $m^s$ gives \begin{align*} 0 = c_m + \sum_{n=m+1}^{\infty} c_n \left(\frac{m}{n}\right)^s. \end{align*} [/step] [step:Show that the rescaled tail tends to zero] Choose a real number $S > \sigma_0$. Since the difference Dirichlet series converges absolutely at $S$, the numerical series \begin{align*} \sum_{n=1}^{\infty} |c_n| n^{-S} \end{align*} converges. For every real $s \geq S$ and every $n > m$, \begin{align*} |c_n| \left(\frac{m}{n}\right)^s = m^S |c_n| n^{-S} \left(\frac{m}{n}\right)^{s-S} \leq m^S |c_n| n^{-S}, \end{align*} because $0 < m/n < 1$. The majorant \begin{align*} \sum_{n=m+1}^{\infty} m^S |c_n| n^{-S} \end{align*} is summable. Let $\varepsilon > 0$. Choose $N > m$ such that \begin{align*} \sum_{n=N+1}^{\infty} m^S |c_n| n^{-S} < \frac{\varepsilon}{2}. \end{align*} For the finite set $\{m+1,\dots,N\}$, each factor $(m/n)^s$ tends to $0$ as $s \to \infty$. Hence there exists $S_\varepsilon \geq S$ such that, for every real $s \geq S_\varepsilon$, \begin{align*} \sum_{n=m+1}^{N} |c_n| \left(\frac{m}{n}\right)^s < \frac{\varepsilon}{2}. \end{align*} Combining the finite part and the tail estimate, for every real $s \geq S_\varepsilon$, \begin{align*} \left| \sum_{n=m+1}^{\infty} c_n \left(\frac{m}{n}\right)^s \right| \leq \sum_{n=m+1}^{N} |c_n| \left(\frac{m}{n}\right)^s + \sum_{n=N+1}^{\infty} |c_n| \left(\frac{m}{n}\right)^s < \varepsilon. \end{align*} Therefore \begin{align*} \lim_{s \to \infty} \sum_{n=m+1}^{\infty} c_n \left(\frac{m}{n}\right)^s = 0, \end{align*} where $s$ tends to infinity through real values. [/step] [step:Take the limit and obtain the contradiction] For every real $s > \sigma_0$, \begin{align*} 0 = c_m + \sum_{n=m+1}^{\infty} c_n \left(\frac{m}{n}\right)^s. \end{align*} Taking the limit as $s \to \infty$ through real values and using the tail limit proved above yields \begin{align*} 0 = c_m. \end{align*} This contradicts the choice of $m$ with $c_m \neq 0$. Hence no such first nonzero coefficient exists, so $c_n = 0$ for every $n \in \mathbb{N}$. Since $c_n = a_n - b_n$, it follows that $a_n = b_n$ for every $n \in \mathbb{N}$. [/step]