Condition on $(a_n)_{n \in \mathbf{N}}$ for Dirichlet-series $\sum_{n = 1}^\infty \frac{a_n}{n^s}$ to converge for $Re(s) > 1$. I'm curious about the conditions on a sequence $(a_n)_{n \in \mathbf{N}}$ of real numbers such that the Dirichlet-series $\sum_{n = 1}^\infty \frac{a_n}{n^s}$ converges absolutely for $Re(s) > 1$. Clearly, this is the case if $a_n$ is bounded. 
But what happens for instance if we let $a_n$ be the number of proper representations of $n$ by a given quadratic form? More generally, which condition on $a_n$ must be given in order that the Dirichlet-series $\sum_{n = 1}^\infty \frac{a_n}{n^s}$ converges absolutely? 
 A: For Dirichlet series in general, we have the following (From Multiplicative Number Theory, I. Classical Theory by Montgomery and Vaughan): 
1. Suppose $\sum \frac{a_n}{n^s}$ converges for some $s_0\in\mathbb{C}$, there is a number $\sigma_c \in \mathbb{R}\cup \{-\infty\}$ such that 
$$
\sum \frac{a_n}{n^s} $$
converges if $\mathrm{Re}(s)>\sigma_c$, diverges if $\mathrm{Re}(s)<\sigma_c$. We say that the Dirichlet series $\sum \frac{a_n}{n^s}$ has an abscissa of convergence $\sigma_c$. 
2. Suppose $\sum \frac{a_n}{n^s}$ converges absolutely for some $s_0\in\mathbb{C}$, there is a number $\sigma_a \in \mathbb{R}\cup \{-\infty\}$ such that 
$$
\sum \frac{a_n}{n^s}$$
converges absolutely if $\mathrm{Re}(s) > \sigma_a$, does not converge absolutely if $\mathrm{Re}(s) < \sigma_a$. We say that the Dirichlet series has an abscissa of absolute convergence $\sigma_a$. 
3. If $\sigma_c\geq 0$, then 
$$
\sigma_c = \limsup_{x\rightarrow\infty} \frac{\log |A(x)|}{\log x}
$$
Here, $A(x)=\sum_{n\leq x} a_n$. This might be the one you looked for. 
4. 
$$
\sigma_c\leq \sigma_a\leq \sigma_c+1.
$$
From here, reference is this.
For Epstein zeta function, consider a positive definite quadratic form $Q$, and by your comment, 
$$
Z_Q(s)=\zeta(2s) \sum \frac{r_Q(n)}{n^s}.
$$
Denote by $R_Q(n)$ the number of ways representing $n$ by $Q$. Let $\Delta$ be the discriminant of $Q$. Then there are finitely many equivalence classes of quadratic forms with discriminant $\Delta$, say $Q_1$, $Q_2$, $\ldots$, $Q_{f}$. Let $Q$ be equivalent to $Q_1$. Then we have 
$$
r_{Q}(n)=r_{Q_1}(n) \leq \sum r_{Q_i}(n) \leq \sum R_{Q_i}(n) = w\sum_{d|n} \left(\frac{\Delta}{d}\right).$$
The expression on the far right, is a Dirichlet convolution of a constant function $f(n)=1$ and $g(n)=\left(\frac{\Delta}{d}\right)$. Since the Dirichlet series associated with $f(n)$ (The Riemann zeta function), and that with $g(n)$ (The Dirichlet L-function) are absolutely convergent for $\mathrm{Re}(s)>1$, we see that the Dirichlet series associated with $r_Q(n)$ is also absolutely convergent  for $\mathrm{Re}(s)>1$. 
For easier proof that the Epstein zeta function is absolutely convergent for $\mathrm{Re}(s)>1$, we use the following:
$$
Q(x,y) = a|x + yz|^2
$$
for some complex $z \in \mathbb{C}-\mathbb{R}$. 
Then by AM-GM inequality,
$$
Q(x,y) \geq c |x||y|$$
for some positive real $c$ depending only on $Q$ and $z$. 
Treat the cases with one of $x$, $y$ being $0$, and none of $x$, $y$ being $0$ seperately. 
