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In the answers to this question, it is established that $$\sum_{n\leq x}\frac{d(n)}{n}=\frac{1}{2}(\log(x))^{2}+2\gamma\log (x)+\gamma^{2}-2\gamma_{1}+O\left(x^{-1/2}\right).$$ A related result can be found on p. 13 of the following paper by Maxie Schmidt: $$\sum_{n \leq x} \frac{d(n)}{n} = \frac{1}{2} (\log(x))^{2} + 2 \gamma \log(x) + O\left(x^{-2/3}\right). $$

While working on the evaluation of a multiple rational zeta series, the following sum came up: $$S_{x} := \sum_{n \leq x} \frac{d(n)}{n-1}. $$

I tried finding the partial sum by means of the hyperbola method, but I haven't succeeded in applying it yet.

Question: can an asymptotic expansion for $S_{x}$ be obtained? If so, how? Are results on this partial sum already present in the literature?

Added: I should mention that I am most interested in such expansions that include the explicit constant terms

Max Lonysa Muller
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1 Answers1

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It is known that

$$ A(x)=\sum_{n\le x}d(n)=x\log x+(2\gamma-1)x+\mathcal O(\sqrt x) $$

Consequently, by a substitution we have

$$ \sum_{2\le n\le x}{d(n)\over n-1}=-1+\sum_{n\le x}{d(n)\over n}+\sum_{2\le n\le x}d(n)\left[{1\over n-1}-\frac1n\right] $$

For the latter sum, we have

\begin{aligned} \sum_{2\le n\le x}d(n)\left[{1\over n-1}-\frac1n\right] &\ll\sum_{2\le n\le x}{d(n)\over n^2}<\zeta^2(2) \end{aligned}

Using the fact that $d(n)\ll_\delta n^\delta$ for every $0<\delta<1$, we have

$$ \sum_{n>x}{d(n)\over n(n-1)}\ll_\delta\sum_{n>x}{1\over n^{2-\delta}}\ll_\delta\int_x^\infty{\mathrm dt\over n^{2-\delta}}\ll_\delta x^{\delta-1} $$

Combining all these together

$$ \sum_{n\le x}{d(n)\over n-1}=\frac12\log^2x+2\gamma\log x+C+\mathcal O(x^{-1/2}) $$

where

$$ C=\gamma^2-2\gamma_1-1+\sum_{n=2}^\infty{d(n)\over n(n-1)} $$

Hope this can help you.

TravorLZH
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