[proofplan] The proof runs Abel summation (summation by parts) on the partial sums of the Dirichlet series. Writing $T(N) = \sum_{n \leq N} a_n$, we decompose $\sum_{n \leq N} a_n n^{-s}$ as a sum of telescoping differences $T(n)(n^{-s} - (n+1)^{-s})$ plus a boundary term $T(N)/N^s$. The growth hypothesis $T(N) = O(N^r)$ makes the boundary term vanish for $\operatorname{Re}(s) > r$, and the main sum is majorised by $\int_1^\infty x^{r - s - 1}\, d\mathcal{L}^1(x)$ (via the mean-value estimate $n^{-s} - (n+1)^{-s} = \int_n^{n+1} s x^{-s-1} \, d\mathcal{L}^1(x)$), which converges there. Uniform convergence on compact subsets of $\{\operatorname{Re}(s) > r\}$ follows by the same estimates with strict inequalities, and holomorphicity is a consequence of **Weierstrass's theorem** on uniform limits of holomorphic functions. [/proofplan]

[step:Set up partial sums and perform Abel summation to rewrite the Dirichlet partial sum]Fix $s \in \mathbb{C}$ with $\operatorname{Re}(s) > r$, and write $s = \sigma + i t$ with $\sigma, t \in \mathbb{R}$ and $\sigma > r$. Define the partial-sum function \begin{align*} T: \mathbb{Z}_{\geq 0} &\to \mathbb{C} \\ N &\mapsto \sum_{n=1}^N a_n, \end{align*} with the convention $T(0) = 0$. Then $a_n = T(n) - T(n-1)$ for $n \geq 1$. Substituting into the partial Dirichlet sum and reindexing the second sum via $m = n - 1$: \begin{align*} \sum_{n=1}^N a_n n^{-s} &= \sum_{n=1}^N \bigl(T(n) - T(n-1)\bigr) n^{-s} \\ &= \sum_{n=1}^N T(n) n^{-s} - \sum_{n=1}^N T(n-1) n^{-s} \\ &= \sum_{n=1}^N T(n) n^{-s} - \sum_{m=0}^{N-1} T(m) (m+1)^{-s}. \end{align*} Since $T(0) = 0$, the $m = 0$ term in the second sum vanishes, so we can start that sum at $m = 1$. Combining the two sums over the common range $1 \leq n \leq N - 1$ and separating the boundary term at $n = N$: \begin{align*} \sum_{n=1}^N a_n n^{-s} = \sum_{n=1}^{N-1} T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr) + T(N) N^{-s}. \end{align*} This is **Abel summation**.[/step]

[guided]Abel summation — the discrete analogue of integration by parts — is the standard tool for converting a sum $\sum a_n b_n$ with known partial sums $T(n) = \sum_{k \leq n} a_k$ into a sum involving differences $b_n - b_{n+1}$. It is powerful here because the hypothesis controls $T(n)$ directly. **Setting up.** Define the partial-sum sequence $T(N) = \sum_{n=1}^N a_n$, with $T(0) = 0$ by convention. The individual term $a_n$ is recovered by the telescoping difference \begin{align*} a_n = T(n) - T(n-1), \quad n \geq 1. \end{align*} The decomposition $s = \sigma + i t$ with $\sigma = \operatorname{Re}(s) > r$ will be used to bound $|n^{-s}| = n^{-\sigma}$. **Substituting and splitting.** Substituting $a_n = T(n) - T(n-1)$ into $\sum_{n=1}^N a_n n^{-s}$: \begin{align*} \sum_{n=1}^N a_n n^{-s} = \sum_{n=1}^N T(n) n^{-s} - \sum_{n=1}^N T(n-1) n^{-s}. \end{align*} Reindex the second sum by $m = n - 1$ (so $n = m+1$, and $n \in \{1, \ldots, N\}$ becomes $m \in \{0, \ldots, N-1\}$): \begin{align*} \sum_{n=1}^N T(n-1) n^{-s} = \sum_{m=0}^{N-1} T(m) (m+1)^{-s}. \end{align*} The $m = 0$ term is $T(0) \cdot 1^{-s} = 0$, so we can start the sum at $m = 1$. Renaming $m$ to $n$: \begin{align*} \sum_{n=1}^N T(n-1) n^{-s} = \sum_{n=1}^{N-1} T(n) (n+1)^{-s}. \end{align*} **Combining.** Subtracting and peeling off the boundary term $n = N$ from the first sum: \begin{align*} \sum_{n=1}^N a_n n^{-s} &= \sum_{n=1}^{N} T(n) n^{-s} - \sum_{n=1}^{N-1} T(n) (n+1)^{-s} \\ &= \sum_{n=1}^{N-1} T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr) + T(N) N^{-s}. \end{align*} This is the Abel summation formula. The left-hand side is the partial Dirichlet sum; the right-hand side expresses it as a main term (sum of $T(n)$ times the "derivative" $n^{-s} - (n+1)^{-s}$) plus a boundary term $T(N) N^{-s}$. Intuitively, we have moved the analytic structure $n^{-s}$ off of the unknown $a_n$ and onto the known differences.[/guided]

[step:Show the boundary term $T(N) N^{-s}$ vanishes as $N \to \infty$ for $\operatorname{Re}(s) > r$]By hypothesis, there exists a constant $B > 0$ such that $|T(N)| \leq B N^r$ for all $N \geq 1$. Since $|N^{-s}| = N^{-\sigma}$ with $\sigma = \operatorname{Re}(s) > r$: \begin{align*} |T(N) N^{-s}| \leq B N^r \cdot N^{-\sigma} = B N^{-(\sigma - r)}. \end{align*} The exponent $\sigma - r > 0$, so $N^{-(\sigma - r)} \to 0$ as $N \to \infty$, and hence $T(N) N^{-s} \to 0$.[/step]

[guided]The boundary term $T(N) N^{-s}$ appears in the Abel summation formula and must vanish in the limit if the Dirichlet series is to equal the main sum alone. We verify this directly from the growth hypothesis. **Extracting a constant from $T(N) = O(N^r)$.** The asymptotic notation $T(N) = O(N^r)$ means there exists $B > 0$ and $N_0$ such that $|T(N)| \leq B N^r$ for all $N \geq N_0$. By enlarging $B$ if necessary to accommodate the finitely many values $N < N_0$, we may assume $|T(N)| \leq B N^r$ for all $N \geq 1$. **Modulus of $N^{-s}$.** For $s = \sigma + it$ with $\sigma \in \mathbb{R}$ and $t \in \mathbb{R}$: \begin{align*} |N^{-s}| = |e^{-s \log N}| = |e^{-\sigma \log N - i t \log N}| = e^{-\sigma \log N} |e^{-i t \log N}| = N^{-\sigma}, \end{align*} since $|e^{i \theta}| = 1$ for real $\theta$. So $|N^{-s}|$ depends only on the real part $\sigma$ of $s$. **Combining the estimates.** \begin{align*} |T(N) N^{-s}| = |T(N)| \cdot |N^{-s}| \leq B N^r \cdot N^{-\sigma} = B N^{-(\sigma - r)}. \end{align*} The hypothesis $\operatorname{Re}(s) > r$ translates to $\sigma - r > 0$, so the exponent of $N$ is strictly negative. Hence $N^{-(\sigma - r)} \to 0$ as $N \to \infty$, and \begin{align*} T(N) N^{-s} \to 0 \quad \text{as } N \to \infty. \end{align*} This is the first place the hypothesis $\operatorname{Re}(s) > r$ is used — it ensures the boundary term vanishes. Without it (say, at $\sigma = r$), the bound would give $|T(N) N^{-s}| \leq B$, which does not guarantee convergence to zero.[/guided]

[step:Estimate the differences $n^{-s} - (n+1)^{-s}$ by an integral using the mean-value identity]For each $n \geq 1$ and each $s$ with $\operatorname{Re}(s) > 0$, the identity \begin{align*} n^{-s} - (n+1)^{-s} = \int_n^{n+1} \frac{s}{x^{s+1}} \, d\mathcal{L}^1(x) \end{align*} holds. This is the Fundamental Theorem of Calculus applied to the function $\varphi: x \mapsto x^{-s}$, which is holomorphic on $\mathbb{C} \setminus \{0\}$ with $\varphi'(x) = -s \cdot x^{-s-1}$ for real $x > 0$, so \begin{align*} n^{-s} - (n+1)^{-s} = -\bigl[\varphi(n+1) - \varphi(n)\bigr] = -\int_n^{n+1} \varphi'(x) \, d\mathcal{L}^1(x) = \int_n^{n+1} \frac{s}{x^{s+1}} \, d\mathcal{L}^1(x). \end{align*} Taking absolute values with $|s/x^{s+1}| = |s| \cdot x^{-(\sigma+1)}$ (since $|x^{-s-1}| = x^{-\sigma - 1}$ for real $x > 0$): \begin{align*} \bigl|n^{-s} - (n+1)^{-s}\bigr| \leq \int_n^{n+1} \frac{|s|}{x^{\sigma + 1}} \, d\mathcal{L}^1(x). \end{align*}[/step]

[guided]We need to bound the telescoping differences $n^{-s} - (n+1)^{-s}$ that appear in the Abel-summation formula. The cleanest approach is the mean-value-type identity obtained by writing the difference as an integral of the derivative over the interval $[n, n+1]$. **Viewing $x \mapsto x^{-s}$ as a smooth function of a real variable.** For real $x > 0$, define \begin{align*} \varphi: (0, \infty) &\to \mathbb{C} \\ x &\mapsto x^{-s} = e^{-s \log x}. \end{align*} Using the chain rule for the composition of the real-variable $\log$ with the entire function $e^{-s \cdot}$: \begin{align*} \varphi'(x) = e^{-s \log x} \cdot (-s) \cdot \frac{1}{x} = -s \cdot x^{-s} \cdot x^{-1} = -s \cdot x^{-s-1}. \end{align*} **Fundamental Theorem of Calculus.** The function $\varphi$ is continuously differentiable on $[n, n+1]$ (any closed interval in $(0, \infty)$), so \begin{align*} \varphi(n+1) - \varphi(n) = \int_n^{n+1} \varphi'(x) \, d\mathcal{L}^1(x) = -s \int_n^{n+1} x^{-s-1} \, d\mathcal{L}^1(x). \end{align*} Negating both sides: \begin{align*} n^{-s} - (n+1)^{-s} = \varphi(n) - \varphi(n+1) = s \int_n^{n+1} x^{-s-1} \, d\mathcal{L}^1(x). \end{align*} **Taking moduli.** For real $x > 0$, $|x^{-s-1}| = x^{-(\sigma+1)}$ by the same computation as in Step 2. Applying the triangle inequality for integrals (valid for continuous integrands on a compact interval): \begin{align*} \bigl|n^{-s} - (n+1)^{-s}\bigr| = \left|s \int_n^{n+1} x^{-s-1} \, d\mathcal{L}^1(x)\right| \leq |s| \int_n^{n+1} x^{-(\sigma+1)} \, d\mathcal{L}^1(x). \end{align*} This replaces the discrete difference with an integral, which is easier to sum against $n^r$ — in Step 4 we will combine it with the bound on $T(n)$.[/guided]

[step:Sum the bounds to deduce absolute convergence of the Abel-transformed series for $\operatorname{Re}(s) > r$]For each $n \geq 1$ and each $x \in [n, n+1]$, the hypothesis $|T(n)| \leq B n^r$ combined with $n \leq x$ gives \begin{align*} |T(n)| \leq B n^r. \end{align*} If $r \geq 0$, then $n^r \leq x^r$ on $[n, n+1]$ (monotonicity of $x \mapsto x^r$). If $r < 0$, then $n^r \leq 2^{|r|} x^r$ on $[n, n+1]$, since $x/n \leq 2$ for $x \in [n, n+1]$ and $n \geq 1$. In either case, there is a constant $C_r \geq 1$ (with $C_r = 1$ for $r \geq 0$ and $C_r = 2^{|r|}$ for $r < 0$) such that \begin{align*} |T(n)| \leq C_r B x^r \quad \text{for all } x \in [n, n+1]. \end{align*} Combining with the integral bound from Step 3: \begin{align*} |T(n)| \cdot \bigl|n^{-s} - (n+1)^{-s}\bigr| &\leq C_r B x^r \cdot |s| \int_n^{n+1} x^{-(\sigma+1)} \, d\mathcal{L}^1(x) \end{align*} is not quite right because we need to put the $x^r$ bound inside the integral. More carefully: since $|T(n)| \leq C_r B x^r$ uniformly for $x \in [n, n+1]$, \begin{align*} |T(n) \cdot \bigl(n^{-s} - (n+1)^{-s}\bigr)| &\leq |s| \int_n^{n+1} |T(n)| \cdot x^{-(\sigma+1)} \, d\mathcal{L}^1(x) \\ &\leq C_r B |s| \int_n^{n+1} x^r \cdot x^{-(\sigma+1)} \, d\mathcal{L}^1(x) \\ &= C_r B |s| \int_n^{n+1} x^{r - \sigma - 1} \, d\mathcal{L}^1(x). \end{align*} Summing over $n \geq 1$: \begin{align*} \sum_{n=1}^\infty \bigl|T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr)\bigr| \leq C_r B |s| \int_1^\infty x^{r - \sigma - 1} \, d\mathcal{L}^1(x). \end{align*} The improper integral on the right is \begin{align*} \int_1^\infty x^{r - \sigma - 1} \, d\mathcal{L}^1(x) = \left[\frac{x^{r - \sigma}}{r - \sigma}\right]_1^\infty = \frac{1}{\sigma - r}, \end{align*} which is finite because $\sigma - r > 0$. Hence the series $\sum_n T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr)$ converges absolutely. Combining with Step 2, the Abel-summation formula of Step 1 gives \begin{align*} \sum_{n=1}^\infty a_n n^{-s} = \lim_{N \to \infty} \sum_{n=1}^N a_n n^{-s} = \sum_{n=1}^\infty T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr) + 0, \end{align*} so the Dirichlet series $\sum a_n n^{-s}$ converges at each $s$ with $\operatorname{Re}(s) > r$, and equals the convergent sum on the right.[/step]

[guided]We now combine the ingredients: Step 2 handled the boundary term, Step 3 provided an integral bound for the differences $n^{-s} - (n+1)^{-s}$, and the hypothesis $|T(n)| \leq B n^r$ controls the coefficients. We assemble these into absolute convergence of the Abel-transformed sum. **Matching $T(n)$ to an $x$-dependent bound.** The difficulty is that $T(n)$ is indexed by $n$, but the integral bound has $x \in [n, n+1]$. We use monotonicity: on the interval $[n, n+1]$, - If $r \geq 0$, then $n^r \leq x^r$ (the function $x \mapsto x^r$ is increasing on $(0, \infty)$ for $r \geq 0$). - If $r < 0$, then $x^r$ is decreasing, so $n^r \geq x^r$. But $x/n \leq 2$ (since $x \leq n+1 \leq 2n$ for $n \geq 1$), so $n^r = (x \cdot n/x)^r = x^r \cdot (n/x)^r$, and $(n/x)^r \leq (1/2)^r \cdot (\text{something} \leq 1)^{-1}$... cleaner: $n/x \geq 1/2$, so if $r < 0$, $(n/x)^r \leq (1/2)^r = 2^{|r|}$. Thus $n^r \leq 2^{|r|} x^r$. In both cases, with $C_r = \max(1, 2^{|r|}) = 2^{\max(0, -r)}$: \begin{align*} n^r \leq C_r \cdot x^r \quad \text{for all } x \in [n, n+1], \ n \geq 1. \end{align*} So $|T(n)| \leq B n^r \leq C_r B x^r$ uniformly in $x \in [n, n+1]$. **Combining with the integral bound.** For each $n \geq 1$: \begin{align*} |T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr)| &= |T(n)| \cdot \bigl|n^{-s} - (n+1)^{-s}\bigr| \\ &\leq |T(n)| \cdot |s| \int_n^{n+1} x^{-(\sigma+1)} \, d\mathcal{L}^1(x) \\ &= |s| \int_n^{n+1} |T(n)| \cdot x^{-(\sigma+1)} \, d\mathcal{L}^1(x) \\ &\leq |s| \int_n^{n+1} C_r B x^r \cdot x^{-(\sigma+1)} \, d\mathcal{L}^1(x) \\ &= C_r B |s| \int_n^{n+1} x^{r - \sigma - 1} \, d\mathcal{L}^1(x). \end{align*} Here the third step pulls $|T(n)|$ inside the integral — valid because $|T(n)|$ is a constant in $x$ — and the fourth step substitutes the uniform bound. **Summing over $n$.** The intervals $[n, n+1]$ for $n = 1, 2, 3, \ldots$ partition $[1, \infty)$ (up to overlap at integer endpoints, which have $\mathcal{L}^1$-measure zero). So \begin{align*} \sum_{n=1}^\infty \bigl|T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr)\bigr| &\leq C_r B |s| \sum_{n=1}^\infty \int_n^{n+1} x^{r - \sigma - 1} \, d\mathcal{L}^1(x) \\ &= C_r B |s| \int_1^\infty x^{r - \sigma - 1} \, d\mathcal{L}^1(x), \end{align*} by monotone convergence of partial sums to the full integral (the integrand $x^{r - \sigma - 1}$ is non-negative, so summing the pieces over disjoint intervals gives the integral over their union). **Evaluating the improper integral.** The exponent $r - \sigma - 1$ is strictly less than $-1$ (since $\sigma > r$ implies $r - \sigma < 0$, so $r - \sigma - 1 < -1$). Hence by elementary calculus, \begin{align*} \int_1^\infty x^{r - \sigma - 1} \, d\mathcal{L}^1(x) = \left[\frac{x^{r - \sigma}}{r - \sigma}\right]_1^\infty = 0 - \frac{1}{r - \sigma} = \frac{1}{\sigma - r} < \infty. \end{align*} **Absolute convergence.** We have shown \begin{align*} \sum_{n=1}^\infty \bigl|T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr)\bigr| \leq \frac{C_r B |s|}{\sigma - r} < \infty, \end{align*} so $\sum_{n=1}^\infty T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr)$ converges absolutely. **Passing to the limit in Abel summation.** From Step 1: \begin{align*} \sum_{n=1}^N a_n n^{-s} = \sum_{n=1}^{N-1} T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr) + T(N) N^{-s}. \end{align*} The main sum converges absolutely as $N \to \infty$ (just shown); the boundary term vanishes (Step 2). Hence the left side has a limit: \begin{align*} \sum_{n=1}^\infty a_n n^{-s} = \sum_{n=1}^\infty T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr), \end{align*} and the Dirichlet series converges at $s$.[/guided]

[step:Upgrade pointwise convergence to uniform convergence on compact subsets of $\{\operatorname{Re}(s) > r\}$ and conclude holomorphicity]Let $K \subseteq \{s \in \mathbb{C} : \operatorname{Re}(s) > r\}$ be compact. Define \begin{align*} \sigma_K := \min_{s \in K} \operatorname{Re}(s), \qquad S_K := \max_{s \in K} |s|. \end{align*} Both are attained because $K$ is compact and $s \mapsto \operatorname{Re}(s)$ and $s \mapsto |s|$ are continuous, and $\sigma_K > r$ because every $s \in K$ satisfies $\operatorname{Re}(s) > r$ and the infimum over a compact subset of the open half-plane $\{\operatorname{Re}(s) > r\}$ is strict. For each $s \in K$, the estimates of Step 4 give \begin{align*} \bigl|T(n) \bigl(n^{-s} - (n+1)^{-s}\bigr)\bigr| \leq C_r B \cdot S_K \int_n^{n+1} x^{r - \sigma_K - 1} \, d\mathcal{L}^1(x), \end{align*} using $|s| \leq S_K$ and $\sigma = \operatorname{Re}(s) \geq \sigma_K$ (which makes $x^{-(\sigma+1)} \leq x^{-(\sigma_K + 1)}$ for $x \geq 1$). The right-hand side is independent of $s \in K$, and sums to a finite quantity: \begin{align*} \sum_{n=1}^\infty C_r B \cdot S_K \int_n^{n+1} x^{r - \sigma_K - 1} \, d\mathcal{L}^1(x) = \frac{C_r B \cdot S_K}{\sigma_K - r}. \end{align*} By the Weierstrass M-test, the series $\sum_{n=1}^\infty T(n)(n^{-s} - (n+1)^{-s})$ converges uniformly on $K$. Similarly, $T(N) N^{-s} \to 0$ uniformly on $K$ (using $|T(N) N^{-s}| \leq B N^{-(\sigma_K - r)}$, which goes to zero independently of $s \in K$). Each partial sum $\sum_{n=1}^{N-1} T(n)(n^{-s} - (n+1)^{-s}) + T(N) N^{-s}$ is a finite sum of entire functions (each $n^{-s} = e^{-s \log n}$ is entire in $s$), hence entire. The uniform limit on every compact subset of $\{\operatorname{Re}(s) > r\}$ is holomorphic by Weierstrass's theorem on uniform convergence of holomorphic functions. Therefore $\sum_{n=1}^\infty a_n n^{-s}$ defines a holomorphic function on $\{\operatorname{Re}(s) > r\}$.[/step]

[guided]Pointwise convergence (Step 4) is not enough: we need holomorphicity, which requires uniform convergence on compact sets. This is where **Weierstrass's theorem on uniform limits of holomorphic functions** enters: if $f_N \to f$ uniformly on compact subsets of an open set $U \subseteq \mathbb{C}$, and each $f_N$ is holomorphic on $U$, then $f$ is holomorphic on $U$. **Extracting uniform bounds on a compact $K$.** Let $K$ be a compact subset of $\{\operatorname{Re}(s) > r\}$. The real-part function $s \mapsto \operatorname{Re}(s)$ is continuous, so on the compact $K$ it attains its infimum: there exists $s^* \in K$ with $\operatorname{Re}(s^*) = \sigma_K := \inf_K \operatorname{Re}$. Since $s^* \in K$ and $K \subseteq \{\operatorname{Re}(s) > r\}$, we have $\sigma_K > r$. Similarly, $s \mapsto |s|$ is continuous and attains its supremum $S_K := \sup_K |s| < \infty$ on $K$. **Uniform bound on the summands.** For any $s \in K$, the estimates of Step 4 become uniform once we replace $\sigma = \operatorname{Re}(s)$ by the worse value $\sigma_K$ (smaller exponent $\sigma$ means larger $x^{-\sigma-1}$) and $|s|$ by the worse value $S_K$: \begin{align*} \bigl|T(n)\bigl(n^{-s} - (n+1)^{-s}\bigr)\bigr| &\leq |s| \int_n^{n+1} |T(n)| \cdot x^{-(\sigma+1)} \, d\mathcal{L}^1(x) \\ &\leq S_K \int_n^{n+1} C_r B x^r \cdot x^{-(\sigma_K + 1)} \, d\mathcal{L}^1(x) \\ &= C_r B S_K \int_n^{n+1} x^{r - \sigma_K - 1} \, d\mathcal{L}^1(x) \\ &=: M_n. \end{align*} The quantity $M_n$ is independent of $s \in K$, as required for the M-test. **Summability of the majorant sequence.** As in Step 4, \begin{align*} \sum_{n=1}^\infty M_n = C_r B S_K \int_1^\infty x^{r - \sigma_K - 1} \, d\mathcal{L}^1(x) = \frac{C_r B S_K}{\sigma_K - r} < \infty, \end{align*} because $\sigma_K - r > 0$. **Weierstrass M-test for uniform convergence.** The M-test states: if $|f_n(s)| \leq M_n$ on a set $K$ with $\sum_n M_n < \infty$, then $\sum_n f_n$ converges uniformly on $K$. Applying this with $f_n(s) = T(n)(n^{-s} - (n+1)^{-s})$ and the $M_n$ just defined, the series $\sum_{n=1}^\infty T(n)(n^{-s} - (n+1)^{-s})$ converges uniformly on $K$. **Boundary term vanishes uniformly.** Using $|s| \leq S_K$ is not needed for the boundary term, but uniformity in $\sigma$ is. For any $s \in K$: \begin{align*} |T(N) N^{-s}| \leq B N^r \cdot N^{-\sigma} \leq B N^{-(\sigma_K - r)}. \end{align*} Since $\sigma_K - r > 0$ is independent of $s \in K$, $N^{-(\sigma_K - r)} \to 0$ uniformly in $s \in K$. Hence $T(N) N^{-s} \to 0$ uniformly on $K$. **Summing these up.** The partial sum $F_N(s) := \sum_{n=1}^N a_n n^{-s}$ equals, by Abel summation, \begin{align*} F_N(s) = \sum_{n=1}^{N-1} T(n)(n^{-s} - (n+1)^{-s}) + T(N) N^{-s}. \end{align*} Both parts converge uniformly on $K$, so $F_N(s) \to F(s) := \sum_{n=1}^\infty T(n)(n^{-s} - (n+1)^{-s})$ uniformly on $K$. **Each $F_N$ is entire.** Each $n^{-s} = e^{-s \log n}$ is entire in $s$ (composition of the entire function $w \mapsto e^{-w \log n}$ with the identity), and $F_N$ is a finite sum of entire functions multiplied by the constants $a_n$. Hence $F_N$ is entire, and in particular holomorphic on $\{\operatorname{Re}(s) > r\}$. **Holomorphicity of $F$ by Weierstrass.** Weierstrass's theorem on uniform limits of holomorphic functions (see, e.g., Conway *Functions of One Complex Variable*, Theorem 7.3.2): if $U \subseteq \mathbb{C}$ is open, $(F_N)$ is a sequence of holomorphic functions on $U$, and $F_N \to F$ uniformly on every compact subset of $U$, then $F$ is holomorphic on $U$. We verify the hypotheses: - $U = \{\operatorname{Re}(s) > r\}$ is open in $\mathbb{C}$. - Each $F_N$ is holomorphic on $U$ (just checked). - $F_N \to F$ uniformly on every compact $K \subseteq U$ (just shown). Hence $F$ is holomorphic on $\{\operatorname{Re}(s) > r\}$, which is the claim of the theorem.[/guided]