2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ 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... $\endgroup$
    – sBs
    Nov 19 '14 at 5:11
  • $\begingroup$ @sBs, Have you found any mistake here? $\endgroup$ Nov 19 '14 at 5:14
  • $\begingroup$ where? In your or my post? I can't find any mistake $\endgroup$
    – sBs
    Nov 19 '14 at 5:22
  • $\begingroup$ @sBs, No no, my post $\endgroup$ Nov 19 '14 at 5:24
  • $\begingroup$ 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? $\endgroup$
    – sBs
    Nov 19 '14 at 5:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.