Dirichlet Series and Asymptotic Expansions: $\tilde{f}(s)= \sum_{n=1}^{\infty} f(n) n^{-s}$ Consider the Dirichlet series $\tilde{f}(s)= \sum_{n=1}^{\infty} f(n) n^{-s}$. In the page Zeta Function Regularization
I found a relation among an asymptotic expansion of $\tilde{f}(s)$ and an asymptotic expansion of $F(t)=\sum_{n=1}^{\infty}f(n)e^{-tn}$, that is if
\begin{equation}
F(t)=\frac{a_N}{t^N}+\frac{a_{N-1}}{t^{N-1}}+...
\end{equation}
then
\begin{equation}
\tilde{f}(s)=\frac{a_N}{s-N}+...
\end{equation}
Is it correct? Where could I find a rigorous treatment of such a subject?
Thank you very much in advance for your help.
 A: Just some basic facts. 
Recall the classic integral representation for the Euler gamma function
$$
\Gamma(s)=\int_0^{+\infty}x^{s-1} e^{-x}dx, \quad \Re s>0, \tag1
$$ then by the change of variables $x=nt$, where $n>0$, we obtain
$$
\frac{1}{n^s}\Gamma(s)=\int_0^{+\infty}t^{s-1} e^{-nx}dx \tag2
$$ then formally
$$
\begin{align}
\Gamma(s)\tilde{f}(s)&= \sum_{n=1}^{\infty} f(n)\:n^{-s} \Gamma(s) \\\\
&=\sum_{n=1}^{\infty} f(n)\int_0^{+\infty}t^{s-1}  e^{-nx}dx \\\\
&=\int_0^{+\infty}t^{s-1}  \sum_{n=1}^{\infty} f(n)e^{-nx}dx \\\\
&=\int_0^{+\infty}t^{s-1}  F(t)dt, \quad \Re s>0, \tag3\\\\
\end{align}
$$ and the previous interchange between the sum and the integral will be justified on the set of absolute convergence of the Dirichlet series. 
If you put $\displaystyle F(t):=\frac{a_N}{t^N}$ in $(3)$, then
$$
"\Gamma(s)\tilde{f}(s)=a_N\int_0^{+\infty}t^{s-1}  \frac{1}{t^N}dt=\frac{a_N}{s-N} " \tag4
$$ and you may see that the poles of $\tilde{f}$ in the s-plane correspond to divergent terms in the Laurent series of its exponential regularization $F$.
Here is a very interesting reference for these considerations: Flajolet Sedgewick book
Hoping to have helped you.
