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my question is , is there a sequence so we have the Dirichlet series

$$ \frac{\zeta(s+1/2)}{\zeta(s)}= \sum_{n=1}^{\infty} \frac{a(n)}{n^s} $$

and the second is, given the dirichlet series for the division function

$$ \zeta (s) \zeta(s-a) =\sum_{n=1}^{\infty} \frac{\sigma _{a}(n)}{n^s} $$

for some $ a > 0 $ , is there a closed formula for

$$ \sum_{x \ge n}\sigma _{a} (n) =A(x) $$

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  • $\begingroup$ Try applying Perron's formula to the known Dirichlet generating function. Asymptotics of the average order sums are known and given here. $\endgroup$ – mds May 24 '18 at 13:54
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There is no closed formula for $$\sum_{n\leq x}\sigma_a(n) $$ but plenty of good approximations are known, from summation by parts and Dirichlet's hyperbola method. About your first question, since: $$ \frac{1}{\zeta(s)} = \sum_{n\geq 1}\frac{\mu(n)}{n^s},\qquad \zeta\left(s+\tfrac{1}{2}\right) = \sum_{n\geq 1}\frac{1/\sqrt{n}}{n^s} $$ by taking $$ a(n)=\sum_{d\mid n}\frac{\mu\left(\tfrac{n}{d}\right)}{\sqrt{d}} = \frac{1}{\sqrt{n}}\sum_{d\mid n}\mu(d)\sqrt{d}$$ we get $$ \sum_{n\geq 1}\frac{a(n)}{n^s} = \sum_{n\geq 1}\frac{1}{n^{s+1/2}}\prod_{p\mid n}\left(1-\sqrt{p}\right)=\frac{\zeta\left(s+\tfrac{1}{2}\right)}{\zeta(s)} $$ as wanted, by Dirichlet's convolution.

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