Cesàro averages of Euler-like functions (Q611965)
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| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Cesàro averages of Euler-like functions |
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Cesàro averages of Euler-like functions (English)
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15 December 2010
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The author proves that that for any complex number \(n\) with real part greater than \(1\), and for any Dirichlet character \(\chi\) (including the trivial character of conductor 1), one has \[ \lim_{N\rightarrow \infty}\frac{1}{N}\sum_{m=1}^{N}\prod_{p|m}\left(1+\frac{\chi(p)}{p^{n-1}}\right)=\frac{L(\chi,n)}{L(\chi^{2},2n)} \] and \[ \lim_{N\rightarrow \infty}\frac{1}{N}\sum_{m=1}^{N}\prod_{p|m}\left(1-\frac{\chi(p)}{p^{n-1}}\right)=\frac{1}{L(\chi,n)}. \] The author at one point writes: ``The number of values of \(m\leq N\) divisible by \(k\) is \([N/k]\), approaching \(N/k\) as \(N\rightarrow \infty\).'' This is incorrect, and should be replaced by \([N/k]=N/k + O(1)\). Dealing with the new error term is easily done, although in Theorem 3 the author should further assume the real part of \(s\) is greater than \(1\) to avoid issues with the new error term (and also issues with the trivial character). In the first proof of Theorem 2, it appears the author claims that the events of being not divisible by a prime \(p\) less than or equal to \(N\), as the prime \(p\) varies, are independent. This is incorrect, especially for primes close to \(N\). It appears that this first proof is meant more as a heuristic argument. The second proof of Theorem 2 is correct, after dealing with the error term more carefully.
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Cesaro average
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arithmetic functions
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