All positive integers but \(1\) are divisble by at least one prime. The probability an integer is divisble by \(p\) is \(\frac{1}{p}\); the complement of that is \(1-\frac{1}{p}\) and so the product over all primes \(\displaystyle \prod_{p \ \text{prime}} \left( 1-\frac{1}{p} \right) = 0\). From the Euler product, we thus know \(\zeta(1)\) is divergent.

Assume the harmonic series converges; then it converges absolutely, since all its terms are positive. This is obviously a contradiction, since we can tell it is conditionally convergent from the ratio/root test.

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TopNewestMy proof for divergence requires no calculus.

\[\dfrac {1}{1} + \dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} + \dfrac {1}{5} + \dfrac {1}{6} + \dfrac {1}{7} + \dfrac {1}{8} + \ldots > \dfrac {1}{1} + \dfrac {1}{2} + \dfrac {1}{4} + \dfrac {1}{4} + \dfrac {1}{8} + \dfrac {1}{8} + \dfrac {1}{8} + \dfrac {1}{8} + \ldots\]

\[\dfrac {1}{1} + \dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} + \dfrac {1}{5} + \dfrac {1}{6} + \dfrac {1}{7} + \dfrac {1}{8} + \ldots > \dfrac {1}{1} + \dfrac {1}{2} + \left( \dfrac {1}{4} + \dfrac {1}{4} \right) + \left(\dfrac {1}{8} + \dfrac {1}{8} + \dfrac {1}{8} + \dfrac {1}{8} \right) + \ldots\]

\[\dfrac {1}{1} + \dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} + \dfrac {1}{5} + \dfrac {1}{6} + \dfrac {1}{7} + \dfrac {1}{8} + \ldots > \dfrac {1}{1} + \dfrac {1}{2} + \dfrac {1}{2} + \dfrac {1}{2} + \ldots\]

\[\dfrac {1}{1} + \dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} + \dfrac {1}{5} + \dfrac {1}{6} + \dfrac {1}{7} + \dfrac {1}{8} + \ldots > \infty\]

Hence proven. – Sharky Kesa · 2 years ago

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– Prasun Biswas · 2 years ago

Right! This is the classical proof of divergence of harmonic series using comparison. This proof is also featured in the Wikipedia page on Harmonic series. Hence, not an original proof.Log in to reply

– Sharky Kesa · 2 years ago

Ye, but nonetheless, easy to remember.Log in to reply

– Jake Lai · 2 years ago

Oresme reincarnate!Log in to reply

– Sharky Kesa · 2 years ago

I'm surprised I know his name. Wasn't Nicole Oresme an astronomer, a mathematician and physicist?Log in to reply

– Jake Lai · 2 years ago

Your proof is often attributed to Oresme.Log in to reply

– Sharky Kesa · 2 years ago

OhLog in to reply

Your proof #1 just went over my head since I'm not quite familiar with Euler product and stuff yet. So, I won't comment on that.

Regarding your proof #2, how can you "tell it is conditionally convergent from the ratio/root test" ? Both ratio and root tests can only be used to test either absolute convergence or divergence. In other cases when the criteria for divergence or absolute convergence is not met, the test is inconclusive. When the tests are inconclusive, the sum

canbe convergent (absolute / conditional) or divergent. You cannot say for sure that the series must be conditionally convergent using ratio / root test in such inconclusive cases (as far as I know). I'm saying all this based on the Wikipedia articles on ratio / root tests.Correct me if I'm wrong. – Prasun Biswas · 2 years ago

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