Proof question: the sum of the reciprocals of the primes diverges

[drjohnsonn]

The gist of the approach I took is that∑1/p = log(e^∑1/p) = log(∏e^1/p) and logx→ ∞ as x→∞.
Proof outline: let ∑1/p = s(x). (...SO I can write this easily on tablet) and note that e^s(x) diverges since e^1/p > 1 for any p and the infinite product where every term exceeds 1 is divergent. Then loge^s(x) diverges as logs as x→∞ would. Thus, since log(e^s(x)= s(x), the sum is found to be divergent

 

[micromass]

the infinite product where every term exceeds 1 is divergent.

That's not right.

 

[D H]

the infinite product where every term exceeds 1 is divergent.

That's obviously incorrect since $\Pi_{n=1}^{\infty} \exp(2^{-n}) = e$.

 

[drjohnsonn]

Wow that isvrry wrong indeed. Thank you

 

[drjohnsonn]

Taking the advice i was initally given - by starting with a product represation of the harmonic series has cerainly panned out better. i keptthinking "my method doesn't seem right somewhere..."

 

[mathwonk]

The proof I have seen (in Trygve Nagell's nice book) is not so immediate. One does get quickly from the Euler product representation of the (divergent) harmonic series, that there are infinitely many primes, but it takes a little estimate to get divergence of your series. Here is a nice wikipedia discussion:

http://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes