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Let $a$ and $b$ be positive integers both greater than $1$. I'd like to compute the average value of the summation \begin{eqnarray} \sum_{k = 0}^{\lfloor \log_{a} x \rfloor} \left \lbrace \log_{b} a^{-k} x \right \rbrace, \end{eqnarray} where $\{ \cdot \}$ denotes the fractional part function and $\lfloor \cdot \rfloor$ denotes the floor function. Plotting some values, it seems to grow really slowly, either logarithmically or double logarithmically.

Here is what I understand. The average value of $\{ \log_{b} x \}$ on $[1,\infty)$ is same as that of $\log_{b} x$ on $[1,b]$ by the periodicity of the fractional part, which I calculate to be $\frac{b}{b-1} - \frac{1}{\log b}$. How does this result apply to the terms of the sum since they are of the form $\{\log_b x - k \log_{b} a \}$? Any help would certainly be appreciated!


This is an extension of my previous question, here.

Edit If I'm understanding things correctly, it should be the case that $\{ \log_{b} cx \}$ has average value on $[1,\infty)$ equal to that of $\log_{b} x$ on $[1, b \ \log_{b} c]$, which I calculate as \begin{eqnarray} \frac{1}{b \ \log_{b} c - 1} \int_{1}^{b \ \log_{b} c} \log_{b} x \ dx = -\frac{\log b + (b \log c) (-1 + \log( b \ \log_{b} c ))}{(\log b)( \log b - b \ \log c)}, \end{eqnarray} which reduces to the value I quoted above in the limit as $c \to b$.

There has to be an easier way of estimating this finite sum!

I do have the naive bounds, \begin{eqnarray} 0 \leq \sum_{k = 0}^{\lfloor \log_{a} x \rfloor} \left \lbrace \log_{b} \ a^{-k} x \right \rbrace \leq \left \lfloor \log_{a} x \right \rfloor + 1. \end{eqnarray}

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