Prove that $\lim_{N \rightarrow\infty}{\sum_{n=1}^N{\frac{\lvert\mu(n)\rvert}{n}}}=\infty$ Prove that $\lim_{N \rightarrow\infty}{\sum_{n=1}^N{\frac{\lvert \mu(n)\rvert}{n}}}=\infty$. I try to write out a few terms and observe what will happen. I notice that the series will somehow be related to $\sum_{n=1}^N{\frac{1}{n}}$ but I don't know how to relate them. Can someone guide me?
 A: We make a very crude estimate of the proportion of square-free numbers in the interval $[2^N+1,2^{N+1}]$. 
Note that $\le \frac{1}{4}$ of these numbers are divisible by $4$, and $\le \frac{1}{9}$ are divisible by $9$, and $\le \frac{1}{25}$ are divisible by $25$, and so on. 
There is overlap, but even if we don't take account of that, the proportion of the numbers in the interval divisible by a square $\gt 1$ is 
$$\le \frac{1}{4}+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}\cdots.\tag{1}$$ 
We can find an upper bound for Sum (1), by using $\frac{1}{4}$ plus
$$\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\cdots,$$
 which is less than 
$$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\cdots.\tag{2}$$ 
But Sum (2) is a telescoping series with sum $\frac{1}{2}$. That is because
$\frac{1}{2\cdot 3}=\frac{1}{2}-\frac{1}{3}$, and $\frac{1}{3\cdot 4}=\frac{1}{3}-\frac{1}{4}$, and so on. 
When we add back the $\frac{1}{4}$ that we left out, we find that Sum (1) is $\le \frac{1}{4}+\frac{1}{2}=\frac{3}{4}$. So the proportion of numbers in $[2^N+1,2^{N+1}]$ that are square-free is at least $\frac{1}{4}$.  
Thus, since there are $2^N$ numbers in the interval, and the reciprocal of each is $\ge \frac{1}{2^{N+1}}$, we have 
$$\sum_{2^N+1}^{2^{N+1}}\frac{|\mu(n)|}{n}\ge 2^N\frac{1}{4}\frac{1}{2^{N+1}}=\frac{1}{8}.$$
Thus each of our infinitely many intervals makes a contribution of at least $\frac{1}{8}$ to the sum, and therefore our sum diverges. 
