Let $\alpha(s) = \sum_{n =1}^{\infty} a_n n^{-s}$ be a Dirichlet series and let $A(x) = \sum_{n \leq x} a_n$. In this article we will establish a relationship between $\alpha(s)$ and $A(x)$.
Theorem. Let $\alpha(s)$ and $A(x)$ be as above. If $\sigma > \max(\sigma_c, 0)$, then
\[
\alpha(s) = s \int_{1}^{\infty} A(x) x^{-s-1} \, dx. \tag{1}
\] Moreover, if $\sigma_c \geq 0$, then
\[
\limsup_{x \to \infty} \frac{\log |A(x)|}{\log x} = \sigma_c. \tag{2}
\]
Proof. Note that by Abel’s summation by parts formula we have
\[
\sum_{n \leq x} a_n n^{-s} = A(x)x^{-s} + s \int_{1}^{x} A(u) u^{-s-1} \, du
\] for every $s$. Now let $\phi$ denote the left-hand side of (2). If $\theta > \phi$, then we have $A(x) \ll x^{\theta}$. Suppose $\sigma > \theta$. Then note that $A(x) x^{-s}$ tends to $0$ as $x \to \infty$ and the integral is absolutely convergent. Thus $\alpha(s)$ converges for $\sigma > \theta$ and so we have $\sigma_c \leq \theta$. Because $\theta$ was taken to be arbitrary we conclude that $\sigma_c \leq \phi$. Moreover, the formula (1) holds for every $\sigma > \phi$.
If $\sigma_c < 0$, then $A(x) \ll 1$ as $\alpha(0)$ converges and so $\phi \leq 0$. Thus in this case (1) holds for every $\sigma > 0$. Now suppose that $\sigma_c > 0$. We will now show that $\phi = \sigma_c$ in this case. We already have the inequality $\sigma_c \leq \phi$ so it suffices to show that $\sigma_c \geq \phi$. Let $\sigma > \sigma_c$. Then by Abel’s summation by parts formula we have
\[
A(x) = \left( \sum_{n \leq x} a_n n^{- \sigma} \right) x^{\sigma}- \sigma \int_{1}^{x} \left( \sum_{n \leq u} a_n n^{-\sigma} \right) u^{\sigma- 1} \, du.
\] Since $\alpha(\sigma)$ converges we have $\sum_{n \leq x} a_n n^{- \sigma} \ll 1$ and so we obtain
\[
A(x) \ll x^{\sigma} + \sigma \int_{0} ^{x} u^{\sigma- 1} \, du \ll x^{\sigma},
\] where the implicit constant depends on $\sigma$. This implies that $\phi \leq \sigma$. Because $\sigma$ was taken to be arbitrary we have $\sigma_c \geq \phi$. This completes the proof.
It follows from what we showed above that $A(x) = o(x^{\sigma})$ for every $\sigma > \max(\sigma_c, 0)$. We now show that if the Dirichlet series $\alpha(s)$ converges for $\sigma > 0$, then $A(x) = o(x^{\sigma})$. This is slightly stronger in the case when $\alpha(\sigma_c)$ converges and $\sigma_c > 0$ for then we have $A(x) = o(x^{\sigma_c})$ which does not follow from above.
Theorem. (Kronecker’s lemma) If $\alpha(s)$ converges and $\sigma > 0$, then $A(x) = o(x^{\sigma})$.
Proof. Let us define $B(x) = \sum_{n \leq x} a_n n^{-x}$. It now follows by Abel’s summation by parts formula that
\begin{align*}
A(x) &= B(x) x^{s}- s \int_{1}^{x} B(u) u^{s- 1}\, du \\
&= s \int_{0}^{x} B(x) u^{s- 1} \, du- s \int_{0}^{x} B(u) u^{s- 1} \, du \\
&= s \int_{0}^{x} (B(x)- B(u)) u^{s- 1} \, du.
\end{align*} Let $\delta(x) = \sup_{u, v \geq x} |B(u)- B(v)|$. Then $\delta(x) = o(1)$ since $B(x) \to \alpha(s)$ as $x \to \infty$. Let $0 < y < x$. Then we have
\[
A(x) \ll \int_{0}^{y} u^{\sigma- 1} \, du + \delta(y) \int_{y}^{x} u^{\sigma- 1} \, du \ll y^{\sigma} + \delta(y) x^{\sigma},
\] where the implicit constant depends on $s$ but we don’t care as $s$ is fixed. Taking $y = \sqrt{x}$ it immediately follows that $A(x) = o(x^{\sigma})$ as $y^{\sigma} = o(x^{\sigma})$ and $\delta(\sqrt{x}) = o(1)$.
The Kronecker’s lemma does not hold for $\sigma < 0$. Consider the Dirichlet series
\[
\alpha(s) = \sum_{n = 1}^{\infty} n^{-2} n^{-s} = \zeta(s + 2)
\] which converges for $\sigma > -1$. Note that
\[
A(x) = \sum_{n \leq x} n^{-2}
\] converges to $\zeta(2)$ and $A(x) \neq o(x^{\sigma})$ for any $-1 < \sigma < 0$.