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Category:2000s romantic drama filmsQ:

Proving that $sum_{k=1}^N mu(k) frac{1}{k} leq ln N$

This is what I am trying to prove
Let $N$ be a positive integer. Prove that:
$$
sum_{k=1}^N mu(k) frac{1}{k} leq ln N
$$
Where $mu$ is the Möbius function and $frac{1}{k}$ is defined to be zero if $k$ is negative or zero if $k=0$ or positive otherwise.
Now, I’m thinking that there is a way to transform this into an integral, but it has to be of this form…
$$
int_0^{+infty} xdmu(x) + x dmu(x)+ mu(x)+ mu(x)ln{x} leq C + ln{N}
$$
Or is there another way to go about proving this?
Thanks!

A:

I don’t think that it’s possible to prove the inequality by integrating a series (there is no such thing as a «simple integral» of Möbius function).
But, if I understand correctly, you are only interested in checking the inequality with numerical evaluation.
One can obtain a very precise estimate of the error using Cesàro means :
$$ sum_{k=1}^{N}frac{mu(k)}{k}leq frac{1}{N}left(sum_{k=1}^{N}frac{mu(k)}{sqrt{k}}right)^2leq frac{1}{N}left(sum_{k=1}^{infty}frac{mu(k)}{sqrt{k}}right)^2= frac{1}{N}cdot 2(8/3-pi^2/6)simeq
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