- #1
henpen
- 50
- 0
My question is relatively breif: is it true that
[tex]\displaystyle \lim_{n \rightarrow \infty}(\varphi(n))=\lim_{n \rightarrow \infty}(n) \cdot \prod_{i=1}^{\infty}(1-\frac{1}{p_i})[/tex]
Where [itex]p[/itex] is prime? Pehaps [itex]\varphi(n)[/itex] is too discontinuous to take the limit of, but it would seem that as it increases to infinity the function should tend to infinity, with fewer anomalies.
If this were true,
[tex]\displaystyle \zeta(1)=\frac{1}{ \prod_{i=1}^{\infty}(1-\frac{1}{p_i})}=\frac{1}{\lim_{n \rightarrow \infty}(\frac{\varphi(n)}{n})}=\lim_{n \rightarrow \infty}(\frac{n}{\varphi(n)})[/tex]
[tex]\displaystyle \lim_{n \rightarrow \infty}(\varphi(n))=\lim_{n \rightarrow \infty}(n) \cdot \prod_{i=1}^{\infty}(1-\frac{1}{p_i})[/tex]
Where [itex]p[/itex] is prime? Pehaps [itex]\varphi(n)[/itex] is too discontinuous to take the limit of, but it would seem that as it increases to infinity the function should tend to infinity, with fewer anomalies.
If this were true,
[tex]\displaystyle \zeta(1)=\frac{1}{ \prod_{i=1}^{\infty}(1-\frac{1}{p_i})}=\frac{1}{\lim_{n \rightarrow \infty}(\frac{\varphi(n)}{n})}=\lim_{n \rightarrow \infty}(\frac{n}{\varphi(n)})[/tex]