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I need estimate the following sum:

$\sum_{d=1}^{n}\frac{\mu(d)}{d}\sum_{k=1}^{\lfloor n/d\rfloor}\frac{1}{k}\frac{q^k}{1-q^{-kd}}$, where $q>1$ and $\mu$ is the Möbius function.

To obtain the main term, I need to find the following sum:

$q+\frac{1}{2}q^2+\cdots+\frac{1}{n}q^n$. Though the second problem looks like calculus problem, I still have no good control about it.

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  • $\begingroup$ Not being sure whether it helps (and maybe in fact this was your starting point) but as $n$ tends to $\infty$, the coefficients of the resulting series stabilize, the $m$th coefficient being$$-\sum_{de=m}\frac1d\frac{\varphi(e-1)}{e-1}$$(where $\varphi$ is the Euler's totient function) $\endgroup$ Apr 11, 2015 at 8:32
  • $\begingroup$ (the sum restricted to $e\ne1$) $\endgroup$ Apr 11, 2015 at 8:41

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Mathematica says that the second sum is equal to:

$$ q^{n+1} (-\Phi (q,1,n+1))-\log (1-q) $$

What do you want to know about it? This function is also known as the Lerch Transcendent, a lot of info can be fund in the Wikipedia article.

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  • $\begingroup$ Hi Professor Rivin, I would like to know the main term of the above sum. i.e. write $q+\frac{1}{2}q^2+\cdots+\frac{1}{n}q^n=Cq^n/n+$ error term for some constant $C$ $\endgroup$ Apr 10, 2015 at 17:09
  • $\begingroup$ @DianbinBao Error term with respect to $n?$ to $q?$ $\endgroup$
    – Igor Rivin
    Apr 10, 2015 at 17:35
  • $\begingroup$ In fact, as $q$ goes to infinity (and $n$ is fixed), the main term is obviously $q^n/n,$ so I assume you mean as $n$ goes to infinity? $\endgroup$
    – Igor Rivin
    Apr 10, 2015 at 17:47
  • $\begingroup$ Note that $\frac{q^{n-1}}{n-1}=\frac{n}{(n-1)q}\frac{q^n}{n}$. This has contribution to the main term. So this is nontrivial. The problem is to find the right constant $C$ in the main term. Here $q$ is fixed and $n\rightarrow\infty$ $\endgroup$ Apr 10, 2015 at 19:51
  • $\begingroup$ Ya, I see the trick, the constant $C\rightarrow \sum_{i=0}^{\infty}\frac{1}{q^i}=\frac{1}{1-q^{-1}}$, which is the zeta value at $s=2$ of the finite field of $q$ elements. $\endgroup$ Apr 10, 2015 at 19:57

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