My question is quite simple and elementary.


Let $A(x)=\sum\_{1}^{x}a(n)$ and $\alpha(s)=\sum\_{1}^{\infty}a(n)n^{-s}$. Then, as we know,
$$ A(x)= \int\_{\gamma-i\infty}^{\gamma+i\infty}\frac{\alpha(s)}{s}x^sds$$
for some $\gamma$ such that $\alpha(\gamma)<\infty$.


Now the question comes, first, we denote $R\_i$ $(i=1,2,3,...)$ as these residues of $\alpha(s)s^{-1}x^s$. And $R\_0$ be the biggest residue among them. Then,
$$ A(x)=O(R\_0)? $$
or are there counter examples?



A question on the big-O value of the complex integral especially in the number theory
Here is a counterexample assuming the Riemann Hypothesis. Let $a(n):=\Lambda(n)-1$, where $\Lambda$ is the von Mangoldt function. Then $A(x)=\psi(x)-[x]$, where $\psi$ is the Chebyshev function, and $\alpha(s)=-\frac{\zeta'(s)}{\zeta(s)}-\zeta(s)$, where $\zeta$ is the Riemann zeta function. So the biggest residue of $\alpha(s)s^{-1}x^s$ is of size $x^{1/2}$, yet we know that
$$ A(x)=\Omega(x^{1/2}\log\log\log x). $$
See Section 14.28 in Titchmarsh: The theory of the Riemann zeta-function.


