Recovering information for $\sum_{n \leq x}a(n)$ from $\sum_{n \geq 1}a(n)e^{-nx}$ I am wondering if I could deduce the bound for the partial sums
\[
\sum_{n \leq x}a(n) \ll x^{A}, \quad x \to \infty
\]
from the relation
\[
\sum_{n \geq 1}a(n)e^{-ny} \ll y^{-A}, \quad y \to 0^{+}.
\]
The condition on $a(n)$ is $|a(n)| = 1$ for all $n \geq 1$.
If anything has been already established on this, it could save my time
tremendously (and much more, might make me happy).
You could use the fact that the function
\[
\sum_{n \geq 1}\frac{a(n)}{n^{s}}
\]
is analytic for $\sigma > A$.
It is a routine that via Mellin inversion theorem,
\[
\sum_{n \geq 1}a(n)e^{-ny} = \frac{1}{2 \pi i }\int_{(3)}
\Gamma(s) f(s) y^{-s}ds.
\]
My first question should be, is it ever possible? 
 A: Perhaps a more conventional way to do this is to take $e^{-y}=z$ and study the power series $F(z) := \sum a(n) z^n$ as $z \to 1^-$.
Then I would consult things on "generatingfunctionology" like 
Chapter 5 of
Wilf, Herbert S., Generatingfunctionology, Wellesley, MA: A K Peters (ISBN 1-56881-279-5/hbk). x, 245 p. (2006). ZBL1092.05001.
or the more comprehensive
Flajolet, Philippe; Odlyzko, Andrew, Singularity analysis of generating functions, SIAM J. Discrete Math. 3, No.2, 216-240 (1990). ZBL0712.05004.
Your question is what they call a "$\Sigma$-transfer", I guess.  Relating your $y$ and their $z$ by $y=\log\frac{1}{z}$ we have: as $y \to 0^+$, $z \to 1^-$, $y \sim (1-z)$.  So your asymptotic condition
$$\sum_{n=1}^\infty a(n)e^{-ny} \ll y^{-A}, \quad y \to 0^{+}.$$
becomes
$$\sum_{n=1}^\infty a(n) z^n \ll (1-z)^{-A}, \quad z \to 1^{-}.$$
Note that you cannot hope to get asymptotic information about $a(n)$ from asymptotic information about $F(z)$ as $z \to 1$ unless $1$ is the unique singularity of $F(z)$ on the unit circle.  That is true in many important applications.  
If there are multiple singularities on the circle of cnvergence, you have to take all of them into account to get asymptotic information about $a(n)$.
