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1/2Happy End Timing Meaning Box Office 69% (276,068,839. All Movie Happy End In Korean Movie Download Free Full HD. Category:1999 films Category:2000s thriller drama films Category:Films directed by Kim Ki-duk Category:Korean-language films 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... $$ \int_0^{+\infty} xd\mu(x) + x d\mu(x)+ \mu(x)+ \mu(x)\ln{x} \leq C +\ln{N} $$ 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)^2\leq \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 f678ea9f9e
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