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

1. Jun 13, 2014

### 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

Last edited: Jun 13, 2014
2. Jun 13, 2014

### micromass

That's not right.

3. Jun 13, 2014

### D H

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

4. Jun 13, 2014

### drjohnsonn

Wow that isvrry wrong indeed. Thank you

5. Jun 13, 2014

### 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..."

6. Jun 13, 2014

### mathwonk

Last edited: Jun 13, 2014