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I've been struggling with the following claim without being able to prove it, so your help would be highly appreciated:

Let $\varphi(n)$ be Euler's totient function. Show that there is a constant $0 such that for any natural number $N$, $KN\leq\frac{\varphi(1)}{1}+\frac{\varphi(2)}{2}+...+\frac{\varphi(N)}{N}$.

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    http://en.wikipedia.org/wiki/Euler%27s_totient_function#Growth_of_the_function2012-03-22

1 Answers 1

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We have that

$\sum_{k=1}^{n} \frac{\varphi(k)}{k} \ge \sum_{k=1}^{n} \left(\left[\frac{n}{k}\right]\frac{k}{n}\right)\frac{\varphi(k)}{k}$

$ = \frac{1}{n} \sum_{k=1}^{n} \left[\frac{n}{k}\right]\varphi(k) = \frac{n+1}{2} $

Thus you can choose $K = \frac{1}{2}$.

The last step uses the identity:

$\sum_{k=1}^{n} \left[\frac{n}{k}\right]\varphi(k) = \frac{n(n+1)}{2} $

and the first inequality uses $\left[\frac{n}{k}\right] \le \frac{n}{k}$, where $[x]$ gives the integer part of $x$.

Multiple proofs of the last identity can be found here: Identity involving Euler's totient function: $\sum \limits_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}$

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    @euler'stotient: You are welcome!2012-03-22