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So I am trying to show the following: $\sum_{n \leq x} \frac{\mu(n)}{n} \log{\frac{x}{n}}=O(1) $ so I tried partial summation as following:

Let $A(x)=\sum_{n \leq x} \frac{\mu(n)}{n}$, then we have $\sum_{n \leq x} \frac{\mu(n)}{n} \log{\frac{x}{n}}= \int_1^{x} \frac{A(t)\mathrm dt}{t},$ and $A(t)$ is clearly very small, or $o(1)$ for all $t \in [1,x]$. My question is how to go from here to conclude that the error term is $O(1)$?

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    Related post on MathOverflow: [$\sum_{d\leq x} (\mu(d)/d) \log x/d$: elementary estimates?](http://mathoverflow.net/q/260235)2017-01-22

1 Answers 1

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As you mentioned, we have $\sum_{n\leq x}\frac{\mu(n)}{n} \log\frac{x}{n}=\int_1^x \left(\sum_{m\leq t}\frac{\mu(m)}{m}\right) \frac{dt}{t}.$

You need a particular bound on the inner sum. For instance, since $\sum_{m\leq t}\frac{\mu(m)}{m}\ll (\log t)^{-2}$, the result follows.

Here is cool trick however. Notice that $\sum_{nd\leq x}\frac{\mu(n)}{nd}=\sum_{k\leq x}\frac{1}{k}\sum_{d|k}\mu(d)=1.$ Now, write the first sum as $\sum_{n\leq x}\frac{\mu(n)}{n}\sum_{k\leq\frac{x}{n}}\frac{1}{k}=\sum_{n\leq x}\frac{\mu(n)}{n}\left(\log\left(\frac{x}{n}\right)+\gamma+O\left(\frac{n}{x}\right)\right).$