- #1
Yuqing
- 218
- 0
From what I had read, Euler had originally proved the infinitude of primes through proving his product formula and equating the primorial to the divergent harmonic series.
[tex]\sum^{\infty}_{n=1}\frac{1}{n^{s}} = \prod^{\infty}_{p}\frac{1}{1-p^{-s}}[/tex]
where p ranges through the primes. However, the proofs of this formula that I have seen all seem to rely on the fact that the sum on the left is absolutely convergent for s>1. I am sure proofs of this formula can be easily found on the web so I am not going to reproduce them here, but I found the proof which employs the sieve of Eratosthenes quite fascinating. However, that proof too relies on rearrangement and at the last step, it requires that the
[tex]lim_{m\rightarrow\infty}\sum^{\infty}_{m}\frac{1}{n^{s}} = 0[/tex]
which isn't the case for the harmonic series. So is the proof actually valid for the case s=1 (or s<1) by modern standards of rigor? If not, can somebody point me towards a valid proof? I assume that the result is true even if the proofs I found were invalid as it appears to be quite celebrated.
[tex]\sum^{\infty}_{n=1}\frac{1}{n^{s}} = \prod^{\infty}_{p}\frac{1}{1-p^{-s}}[/tex]
where p ranges through the primes. However, the proofs of this formula that I have seen all seem to rely on the fact that the sum on the left is absolutely convergent for s>1. I am sure proofs of this formula can be easily found on the web so I am not going to reproduce them here, but I found the proof which employs the sieve of Eratosthenes quite fascinating. However, that proof too relies on rearrangement and at the last step, it requires that the
[tex]lim_{m\rightarrow\infty}\sum^{\infty}_{m}\frac{1}{n^{s}} = 0[/tex]
which isn't the case for the harmonic series. So is the proof actually valid for the case s=1 (or s<1) by modern standards of rigor? If not, can somebody point me towards a valid proof? I assume that the result is true even if the proofs I found were invalid as it appears to be quite celebrated.