# 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?

• Something's off here, the bound for the second series decreases as $y$ approaches zero (if $A>0$). It probably should be $y^{-A}$. Also, if $a(n)\ge 0$, this corrected statement is trivial then. – Christian Remling Jun 1 '18 at 17:41
• It is certainly not true without additional restrictions on $a(n)$, like $a(n)\geq 0$, for example (and I assume that you corrected $A$ to $-A$ as Christian suggests.) – Alexandre Eremenko Jun 1 '18 at 17:47
• @ChristianRemling Definitely. I edited on the second relation. – Mr. SnowRemover Jun 2 '18 at 1:02
• @AlexandreEremenko I am interested in the situation with $|a(n)| = 1$; in particular of the form $a(n) = e^{-\alpha i b(n)}$ with $\alpha$ real and $b(n)$ a real-valued arithmetical function. – Mr. SnowRemover Jun 2 '18 at 1:05
• You are looking for a Tauberian type theorem. When one is concerned with asymptotics instead of inequalities, than the Hardy-Littlewood tauberian theorem is very relevant to your question. – Ofir Gorodetsky Jun 2 '18 at 14:52

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 $a(n)$.

• The cited paper seems to be about coefficients. Any relation between the unique singularity and the partial sums? – Mr. SnowRemover Jun 4 '18 at 1:07