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It is easy to show that

$$ \sum_{d|n}{\frac{\mu(d)}{\phi(d)}} = \prod_{p|n}{\frac{p-2}{p-1}} $$ (1)

using the techniques described in this post: Is there a "nice" formula for $\sum_{d|n}\mu(d)\phi(d)$?. However, I have come across the scenario where instead of summing across the divisors of $n$, the sum is across all whole numbers less than $n$

$$ \sum_{k=1}^{n}{\frac{\mu(k)}{\phi(k)}} $$ (2)

The first thought I had approaching this was to consider the divisors of the primorial $n\#$ because of the integers $1,2...n$, only the divisors of a primorial have nonzero Mobius values. However, the sum is taken only up to $n$, leaving out a large number of divisors. These divisors have a much smaller effect on the sum due to their size, but it is not negligible. How should I go about evaluating this sum? (WolframAlpha doesn't have a nice simplification, just values)

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kg583
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    It is as complicated as showing $\sum_{n=1}^\infty \frac{\mu(n)}{n}$ converges. And $\sum_{n=1}^N \frac{\mu(n)}{\phi(n)} \to 0$ but we don't know at which rate. $\sum_{n=1}^N \frac{\mu(n)}{\phi(n)} = \mathcal{O}(N^{-1/2+\epsilon})$ is the Riemann hypothesis. – reuns Nov 26 '16 at 05:36

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