Proof of $\sum_{n=1}^\infty a_n n^{-s} = s\int_1^\infty A(x) x^{-s-1}\, dx$ and $\limsup_{x\to\infty} \frac{\log|A(x)|}{\log x} = \sigma_c$

Question

Theorem. Let $A(x) := \sum_{n\le x} a_n$. If $\sigma_c < 0$, then $A(x)$ is a bounded function, and $$\sum_{n=1}^\infty a_n n^{-s} = s\int_1^\infty A(x) x^{-s-1}\, dx \tag{1}$$ for $\sigma > 0$. If $\sigma_c\ge 0$, then $$\limsup_{x\to\infty} \frac{\log|A(x)|}{\log x} = \sigma_c \tag{2}$$ and $(1)$ holds for $\sigma > \sigma_c$.

I have questions about the proof.

Proof. We note that $$\begin{align*} \sum_{n=1}^N a_n n^{-s} = \int_{1^-}^N x^{-s}\, dA(x) &= A(x) x^{-s}\vert_{1^-}^N - \int_{1^-}^N A(x)\, dx^{-s}\\ &= A(N) N^{-s} + s\int_1^N A(x) x^{-s-1}\, dx. \tag{3} \end{align*}$$ Let $\phi$ denote the left-hand side of $(2)$. If $\theta > \phi$, then $A(x) \ll x^{\theta}$ where the implicit constant may depend on the $\{a_n\}$ and on the $\theta$. Thus, if $\sigma > \theta$, then the integral in $(1)$ is absolutely convergent. Thus we obtain $(1)$ by letting $N \to \infty$, since the first term above tends to $0$ as $N \to\infty$.

1. Why does the first term above tend to $0$ as $N \to\infty$? We do not a priori know that $A(N)$ is bounded.

Suppose that $\sigma_c < 0$. Then, we know that $A(x)$ tends to a finite limit as $x\to\infty$, and hence $\phi \le 0$, so that $(1)$ holds for all $\sigma > 0$.

2. $\sigma_c < 0$ implies convergence for $s = 0$, so $\lim_{x\to\infty} A(x) < \infty.$ How does this give $\phi \le 0$ and $(1)$ for all $\sigma > 0$?

Now suppose that $\sigma_c \ge 0$. Then, we know that the series in $(1)$ diverges when $\sigma < \sigma_c$. Hence, $\phi \ge \sigma_c$.

3. How do we get $\phi \ge \sigma_c$?

To complete the proof it suffices to show that $\phi\le \sigma_c$. Choose $\sigma_0 > \sigma_c$. Then, by $$\sum_{n=M+1}^N a_n n^{-s} = R(M) M^{s_0-s} - R(N) N^{s_0-s} + (s_0 -s)\int_M^N R(u) u^{s_0-s-1}\, du$$ with $s = 0$ and $M = 0$ (where $R(u) = \sum_{n > u} a_n n^{-s_0}$), we obtain $$A(N) = -R(N)N^{\sigma_0} + \sigma_0 \int_0^N R(u) u^{\sigma_0 - 1}\, du.$$ Since $R(u)$ is a bounded function it follows that $A(N) \ll N^{\sigma_0}$, where the implicit constant may depend on the $\{a_n\}$ and on the $\sigma_0$. Hence, $\phi\le \sigma_0$. Since this holds for any $\sigma_0 > \sigma_c$, and thus $\phi\le \sigma_c$.

4. I see $A(N) \ll N^{\sigma_0}$, but not how $\phi\le \sigma_0$ follows.

Thanks for your help!

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Reference: Theorem $1.3$, Multiplicative Number Theory I, by Montgomery and Vaughan.

Answer 1

1. $A(x)\ll x^\theta$ and $\sigma>\theta$ so $\vert A(N)/N^s\vert=\vert A(N)/N^\sigma\vert\ll 1/N^{\sigma-\theta}\to0$.
2. $\limsup_{x\to\infty} \frac{\log|A(x)|}{\log x}\leq\limsup_{x\to\infty} \frac{C}{\log x}=C\limsup_{x\to\infty} \frac{1}{\log x}=0$.
3. As established in $1.$ for $\phi<\sigma$ we have that $(1)$ holds with the integral being absolutely convergent so the sum converges. If $\phi<\sigma_c$ we could get a contradiction.
4. $\limsup_{x\to\infty} \frac{\log|A(x)|}{\log x}\leq\limsup_{x\to\infty} \frac{\log x^{\sigma_0}}{\log x}=\sigma_0$.