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Jun 30, 2013 at 7:10 vote accept CommunityBot
Jun 30, 2013 at 2:47 comment added KConrad Well, strictly speaking, by rearranging terms you'd get $\exp(\sum A_k/k) = \prod_p \exp(\sum 1/kp^k) = \prod 1/(1-1/p)$, which is $\sum 1/n$ by one argument or another.
Jun 30, 2013 at 2:41 comment added KConrad @VivekShende: That's true. Since everything in sight is positive, convergence of $A_1$ would imply convergence of the harmonic series.
Jun 30, 2013 at 2:36 comment added Vivek Shende As far as concluding $\sum 1/p = \infty$ goes, I don't understand what's wrong with Euler's original argument. If $A_1 = \sum 1/p$ was really finite, then nothing would stop us from expanding the finite quantity $\mathrm{exp}(\sum A_k / k)$ and getting $\sum 1/n$, contradiction.
Jun 29, 2013 at 22:55 history answered KConrad CC BY-SA 3.0