On Mobius function

Let $n \geq 3$. Prove that:

$$\sum_{k=1}^{n} \mu(k!)=1$$

Solution

We have that:

• $\mu(1!)=\mu(1)=1$.
• $\mu(2!)=\mu(2)=(-1)^{1}=-1$
• $\mu(3!)=\mu(1\cdot 2\cdot 3)=(-1)^{2}=1$
• $\mu(4!)=\mu(1\cdot 2\cdot 3 \cdot 4)=\mu(2^2 \cdot 3)=0$

Now, we note that for $n \geq 3$ the $2^2 \mid n!$ hence $\mu(n!)=0$. Thus:

$$ \sum_{k=1}^{n} \mu(k!)= 1 -1 +1 +0 +\cdots +0 = 1$$