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Consider the sequence $\lbrace \frac{\phi(i)}{i}\rbrace_{i=1}^\infty$ where $\phi$ is the Euler's function. The Sequence is clearly dense in $[0,1]$. What can be said about the limsup of its average sequence ?? I mean the sequence $\frac 1n\sum_{i=1}^n \frac{\phi(i)}{i}$. Its value or an upper bound would be helpful.

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1 Answer 1

up vote 7 down vote accepted

It is well known that $\sum_{k\le x} \phi(k)=\frac{3}{\pi^2}x^2+O(x\log(x))$. In a similar way we obtain $$ \sum_{k\le x} \frac{\phi(k)}{k}=\frac{6}{\pi^2}x+O(\log(x)), $$ by using the Moebius function, i.e., $\sum_{k\le x} \frac{\phi(k)}{k}=\sum_{k\le x}\sum_{d\mid k}\frac{\mu(d)}{d}=\sum_{d\le x}\frac{\mu(d)}{d}\lfloor \frac{x}{d}\rfloor$. Then $$ \sum_{k\le x} \frac{\phi(k)}{k}=x\sum_{d\le x}\frac{\mu(d)}{d^2}-\sum_{d\le x}\frac{\mu(d)}{d}((x/d))=\frac{6}{\pi^2}x+O(\log(x)). $$

Edit: The term $O(1)$ (which I hade before from the paper of Carella) is not correct, please see the valuable comment of Peter.
The value $6/\pi^2$ is reasonably good for the question posed, if $n$ is large.

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5  
Unfortunately Carella is a well-known crank, and the claimed error term is false. In fact, Montgomery proved that the error term $H(x)$ satisfies $H(x)=\Omega(\sqrt{\log\log x})$. On the other hand, Walfisz proved that $H(x)=O((\log x)^{2/3}(\log\log x)^{4/3}))$. Actually in both cases these bounds were for the error term $R(x)=\sum_{n\le x}{\varphi(n)}-\frac{3x^2}{\pi^2}$, and then both results follow from a result of Chowla showing that $H(x)-\frac{R(x)}{x}=O((\log x)^{-4})$; see mathoverflow.net/questions/126890/…. –  Peter Humphries Jun 21 '13 at 15:42
    
By the way, Carella acknowledges in his paper Hugh Montgomery for "comments and correcting a few errors". Don't know whether Montgomery has ever seen the paper, though ... . –  Stefan Kohl Jun 21 '13 at 17:33
    
@Peter: thank you. This is important to know. –  Dietrich Burde Jun 21 '13 at 20:37

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