# Why is $\sum_{k=1}^{\infty }\frac{\mu (k)}{k\phi (k)} = \prod_p \left( 1- 1/(p(p-1))\right)$?

That's the actual question - why is $\sum_{k=1}^{\infty }\frac{\mu (k)}{k\phi (k)} = \prod_p \left( 1- \frac{1}{p(p-1)}\right)$?

HINT:

For prime $p,$

$$\sum_{k=1}^{\infty }\frac{\mu (k)}{k\phi (k)}=\prod_p\left[\sum_{r=0}^\infty \frac{\mu (p^r)}{p^r\phi (p^r)}\right]$$

Now, $$\sum_{r=0}^\infty\frac{\mu (p^r)}{p^r\phi (p^r)}=1+\frac{(-1)}{p(p-1)}+0$$

• Thanks, that helps a lot, but: If I want to calculate $\sum_{r=0}^{\infty }\frac{(-1)^r}{p^rp^{r-1}(p-1)}=\frac{1}{p(p-1)}\sum_{r=0}^{\infty }\frac{(-1)^r}{p^{2r}}=\frac{1}{p(p-1)}\cdot \frac{p^2}{p^2+1}$, but that is not that what I wanted...
– sBs
Nov 19 '14 at 5:11
• @sBs, Have you found any mistake here? Nov 19 '14 at 5:14
• where? In your or my post? I can't find any mistake
– sBs
Nov 19 '14 at 5:22
• @sBs, No no, my post Nov 19 '14 at 5:24
• Oh yes I did (in my post), first step - but then I got at the end $\frac{p^3}{(p^2+1)(p-1)}$...is that what I wanted?
– sBs
Nov 19 '14 at 5:24