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Are there useful lower and upper (preferably sharp or asymptotic) bounds for the following fractional part-power summation? Given arbitrary reals $a, b > 1$ and integer $n \geqslant 1$, let \begin{align} f_{n}(x) = \sum_{i = 0}^{\lfloor \log_a x \rfloor} \left( \text{frac}( \log_{a} (b^{-i} x) ) \right)^{n}, \end{align} where $\text{frac}$ denotes the fractional part function. Naive bounds include $0 \leqslant f_n(x) \leqslant \cdots \leqslant f_{1}(x) \leqslant f_{0}(x) = \lfloor \log_a x \rfloor + 1$ for $x \geqslant 1$, so $f_{n}(x) = O(\log x)$, but I suspect that this can be improved substantially for certain special values of $a$ and $b$.

As the order of divergence should depend on the algebraic nature of the factor $\log_{a} b$ by Elementary Equidistribution Theory, I expect a more rapid divergence in the limit $x \to \infty$ if $\log_{a} b$ is rational, perhaps $O(\log x)$. My guess is $O(\log \log x)$ or something similar if $\log_a b$ is irrational.

Any help is certainly appreciated!

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Clearly we can write $\mathcal{O}(\log x)$ instead of just $\mathcal{O}(x)$. –  anon Mar 2 '12 at 3:58
    
Can you throw in a few parentheses so I can tell whether you have $(\log_ab)^{-i}x$ or $\log_a(b^{-i}x)$? frac$(u^n)$ or $({\rm frac}(u))^n$? –  Gerry Myerson Mar 2 '12 at 4:59
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