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siriusmart

Infinite product follow-up (links.hackliberty.org)

submitted 6 months ago by [email protected] to c/dailymaths

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This is related to the the May 16 post, but takes only the prime indexed terms. Does it still diverge?

Hint

Transform the product into a sum

Hint

The harmonic series 1 + 1/2 + 1/3 + ... 1/n +... diverges

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[–] CharlesMangione 6 points 6 months ago* (last edited 6 months ago) (1 children)

I don't know how to begin proving it, but the more I run this series out, bigger it gets. The conditions of the equation are such that it will always have a consistently non-zero rate of increase, even though that rate of increase decreases each time the formula is cycled ((p~n~/p~n~-1) will always be more than (p~n+1~/p~n+1~-1), nonetheless any and every (p~n~/p~n~-1) will be >1). The divergence will be glacial, but definite.

[–] [email protected] 4 points 6 months ago

I can confirm that your intuition for divergence is correct.

[–] siriusmart 4 points 6 months ago* (last edited 6 months ago) (1 children)

I've shown that ln(n/n-1) is always larger than 1/n, so Σln(n/n-1) for all natural number n will be larger than the series 1+1/2+1/3+...

but I don't know how to make sure the sum of all ln(p/p-1) only when p is prime is larger than the provided series

the question is strongly suggesting its divergent, i just dont know how to show it

[–] [email protected] 4 points 6 months ago

Perhaps surprisingly, that's actually good enough since the sum of the prime reciprocals also diverges. However, I'm not letting you just assume that, and proving it is harder than the original problem.

[–] [email protected] 4 points 5 months ago* (last edited 5 months ago)

Solution

What I like about this solution is that it doubles as a proof that there are infinitely many prime numbers. It's not circular either -- it uses the number theoretic fact that every integer has a prime factorization, but nothing deeper. If only finitely many primes were required for that, ... well then of course every finite product converges.