One of three of a kind - Part (3) - Final

Great problem! Part 1 and 2 are definitely needed for this problem!

Suppose $\prod _{ k=1 }^{ \infty } \left( 1-\frac { 1 }{ \phi ^{ n } } \right) ^{ \frac { \mu \left( k \right) -\varphi \left( k \right) }{ k } }=S$ We can log both sides to obtain $\ln { S } =\ln { \prod _{ k=1 }^{ \infty } \left( 1-\frac { 1 }{ \phi ^{ n } } \right) ^{ \frac { \mu \left( k \right) -\varphi \left( k \right) }{ k } } } =\sum _{ k=1 }^{ \infty } \frac { \mu \left( k \right) -\varphi \left( k \right) }{ k } \ln { \left( 1-\frac { 1 }{ \phi ^{ n } } \right) } =\sum _{ k=1 }^{ \infty } \frac { \mu \left( k \right) }{ k } \ln { \left( 1-\frac { 1 }{ \phi ^{ n } } \right) } -\sum _{ k=1 }^{ \infty } \frac { \varphi \left( k \right) }{ k } \ln { \left( 1-\frac { 1 }{ \phi ^{ n } } \right) }$

Since $\frac { 1 }{ \phi } <1$ , and using the fact that $\phi =\frac{1+\sqrt{5}}{2}$ , we can derive that $\sum _{ k=1 }^{ \infty } \frac { \mu \left( k \right) }{ k } \ln { \left( 1-\frac { 1 }{ \phi ^{ n } } \right) } -\sum _{ k=1 }^{ \infty } \frac { \varphi \left( k \right) }{ k } \ln { \left( 1-\frac { 1 }{ \phi ^{ n } } \right) } =1$

I'm not going to show why here, since it is what part 1 and 2 of this question is about, and I don't wish to reveal the answer here.

Hence $\ln{S}=1, \quad S=e$