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Let $a$ and $b$ be positive reals both greater than $1$. I'd like to compute the limit of the summation \begin{eqnarray} \lim_{x \to \infty} \frac{1}{\log_a x} \sum_{k = 0}^{\lfloor \log_{a} x \rfloor} \left \lbrace (\log_{b} x) - k \log_b a \right \rbrace, \end{eqnarray} where $\{ \cdot \}$ denotes the fractional part function and $\lfloor \cdot \rfloor$ denotes the floor function. I believe that the limit is $\frac{1}{2}$ for almost all $a$ and $b$, and I believe the limit doesn't exist if $b = a^r$ where $r$ is a positive rational. I also have a feeling that Weyl's theorem may be playing a role here but I'm not sure if $\lbrace (\log_{b} x) - k \log_b a \rbrace$ is equidistributed on any interval.

Any help is certainly appreciated!

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up vote 2 down vote accepted

Your notation is unclear, but I'll assume you meant $$ \log_b(a^{-k}x) = \log_b(x) - k\log_b a. $$ Weyl's theorem shows that if $\log_b a^{-1}$ is irrational then $\{k\log_b a^{-1}\}$ is asymptotically equidistributed. Shifting it by $\{\log_b x\}$ shouldn't make a difference, and so the limit approaches $1/2$.

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Thanks. I added some parentheses to clarify the function. – user02138 Feb 28 '11 at 22:40

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