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If an arithmetic function $f(n)$ has Dirichlet series $\zeta(s) \prod_{i,j = 1} \frac{\zeta(a_i s)}{\zeta(b_j s)}$, for which values of $a_{i}$ and $b_{j}$ is the following true? That \begin{align} \lim_{x \to \infty} \tfrac{1}{x} \sum_{n \leq x} f(n) = \prod_{i,j = 1} \frac{\zeta(a_i )}{\zeta(b_j )} \end{align} or, more generally, there is a $\kappa > 0$ such that \begin{align} \sum_{n \leq x} f(n) = x \prod_{i,j = 1} \frac{\zeta(a_i )}{\zeta(b_j )} + O(x^{1-\kappa}) \sim x \prod_{i,j = 1} \frac{\zeta(a_i )}{\zeta(b_j )}. \end{align} Does the Wiener-Ikehara Theorem apply here?

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    As long as $a_i,b_i>1$, everything works out. I try to make a proof.2012-06-06
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    Ok, here is the result: The asymptotic is true for $a_i,b_i>1$, and the $\kappa$ exponent in the error term can be as large as $$\min_i \{ (a_i -1),(b_i-1), 1\} -\epsilon$$ for any $\epsilon$.2012-06-06
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    Sweet! Thanks, Eric.2012-06-06
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    I imagine one need only expand the zeta function near its singularity and the rest should follow, no?2012-06-06

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The following theorem appears in an introductory chapter in a soon to be published book by Andrew Granville and Kannan Soundararajan. An early version of the book which contains this material can be found on Granville's website. (Note: The statement of the result may appear different, but the proof is in the book which is currently on his website.)

Theorem: Let $f(n)=1*g(n)$ be a multiplicative function, and suppose that for $0\leq \sigma \leq 1$ the sum $$\sum_{d=1}^\infty \frac{|g(d)|}{d^{\sigma}}=\tilde{G}(\sigma)$$ converges. Then, if we write $\mathcal{P}(f)= \sum_{n=1}^\infty \frac{g(n)}{n},$ we have that $$\left|\sum_{n\leq x} f(n)-x\mathcal{P}(f)\right|\leq x^{\sigma}\tilde{G}(\sigma).$$

With your definition of $f(n)$, we see that $\mathcal{P}(f)=\prod_{i,j} \frac{\zeta(a_i)}{\zeta(b_i)}$, and letting $$\delta=\min\{a_i-1,\ b_i-1,\ 1\},$$ we see that by setting for $\sigma=\delta+\frac{1}{\log x}$, we get $$\sum_{n\leq x}f(n)=x\prod_{i,j}\frac{\zeta(a_i)}{\zeta(b_i)}+O\left(x^{1-\delta}\log x\right).$$

Remark: As an almost immediate corollary, since $\frac{\phi(n)}{n}$ has Dirichlet series $\frac{\zeta(s)}{\zeta(s+1)}$, we get that $$\sum_{n\leq x}\frac{\phi(n)}{n}=\frac{x}{\zeta(2)}+O(\log x).$$

See also this answer on Math.Stackexchange.

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    Hi Eric. What is the title (or link) of the book on Granville's website? Thanks in advance.2012-06-17
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    @user02138: The book goes by two names, either Pretentious Number Theory, or Multiplicative Number Theory.2012-06-22
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    [This](http://www.dms.umontreal.ca/~andrew/PDF/PretendBook050111.pdf) one?2012-06-22
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    @user02138: Yes. Unfortunately that is from a year and a half ago, and I am not sure if he will post a new version any time soon. What I mentioned follows from the work done on pages 20 and 21, but it is very hard to see this connection.2012-06-22