As is usually the case when I ask a math question, I am forced to preface it by saying that my question may not really make a lot of sense …
So, IIRC, the sum of the inverse primes diverges to infinity. Is there an infinite sum, whose terms are “less dense” than the inverse primes, that nevertheless still diverges? For example, if you only consider the sum of the inverses of those primes that end in ‘3’, that series would be less dense than the inverse primes, but would it still diverge?
[sub][sup]I guess by “less dense” than the inverse primes I mean that the average difference between successive terms is smaller than the average difference between successive inverse primes.[/sup][/sub]
I believe that the sum of the reciprocals of the primes less than N goes as ln(ln(N)).
If you had a series that went as ln(ln(ln(N))), that would also diverge, and it seems like it would have to be less dense, right?
It’s pretty easy to construct such a series:
s[sub]0[/sub] = 1
a[sub]n[/sub] = 1/x, where x is the smallest integer such that s[sub]n-1[/sub] + 1/x < ln(ln(ln(x)))
s[sub]n[/sub] = s[sub]n-1[/sub] + a[sub]n[/sub]
Now of course, if this worked, this would not be the least dense divergent subset of { 1/n }. You could do one for ln(ln(ln(ln(N)))). It seems to me that there probably isn’t any least dense set. No proof, of course.
Achernar pretty much has it, this is just an alternate way of looking at it.
There is a close relation between sums and integrals, if the terms are “nice”, e.g., uniformly decreasing. So, more or less, the sum diverges if the integral of the (“nice”) function in the terms gives an unbounded function.
Take your favorite slow growing, but unbounded function. Iterative logs will work. There’s also inverse Ackermann’s function (“alpha”) and even the Busy Beaver Function (so slow growing it’s uncomputable.) Take the derivative of that and use it as the basis of your terms. E.g, ln x is unbounded, derivative is 1/x to the sum of 1/x diverges (the harmonic series).
Since there isn’t a slowest growing unbounded “nice” function, there’s no limit as to how “sparse” a series can be and still diverge.
(Explaining what would make a function “nice” here takes a good chunk of a term in Advanced Calculus.)
Come now, you can’t tantalize us like that without giving more information!
And the inverse primes diverge? I’ve often wondered about that. I wrote a computer program once to generate partial sums, and gave up after about a million terms (it had already passed e by then, but not yet pi, which seemed like a semi-reasonable guess for the sum, if it did converge).
I think I may be miss-understanding the use of the term “Least Dense” on the offchance that I am not the only doper doing that, please critique my following question.
Would not a series whose first term is finite, but the next term is infnite be divergent in the “Least Dense” sense.
Sum (n, 0->inf) { 1/(n-1) }
comes to mind? Or does this not count since, after the n=1 term which diverges, there follows a dense set of finite terms?
I seem to recall another proof to demonstrate the divergence of the sum of the inverse primes. Basically you show that the sum of any infinite series, 1/n[sub]i[/sub] , converges iff the infinite product (1 + 1/n[sub]i[/sub]) converges. Once you’ve done that, it’s a matter of showing that the infinite product of the (1 + 1/p[sub]i[/sub]) diverges and that’s fairly straightforward since the expansion of the product will include the sum of the reciprocals of all the natural numbers (and that series diverges).
In terms of the first part of the proof, we have:
(1+1/n[sub]1[/sub])(1+1/n[sub]2[/sub])(1+1/n[sub]3[/sub]) … < (e[sup]1/n[sub]1[/sub][/sup])(e[sup]1/n[sub]2[/sub][/sup])(e[sup]1/n[sub]3[/sub][/sup]) … = e[sup]sum of the 1/n[sub]i[/sub]'s[/sup]
If the sum of the 1/n[sub]i[/sub]'s converges, so must e to the power of the sum of the 1/n[sub]i[/sub]'s converge, and so then must the infinite product.
It’s been almost thirty years since I’ve done math, so please forgive if none of this makes sense or is totally wrong.
I’m gonna give this thread one itsy-bitsy bump. (After my last response, it fell off the front page pretty rapidly and had essentially no views, let alone replies.) In this way, maybe one of the mathematically inclined among you can let me know if the proof that I recalled (above) is legit (or even makes sense).
