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The harmonic series, after removing any term with a "9" in it, is convergent. (hippomath.blogspot.com)
62 points by japaget on July 3, 2011 | hide | past | favorite | 23 comments


Another interesting one is that if you remove everything but the prime numbers, it still diverges. So there are more primes than numbers without a 9 in them.


Given that prime frequency is around ln(n) on average, and that nineless frequency is around .9^log_10(n) (log-exponentially low, or whatever the right terminology is), this isn't actually that surprising when you think about it.


Well of course, the vast majority of large numbers have nines in them. All but 0% of them.


This is either a typo or you meant something that I did not understand---explain?


Large numbers tend to have a large number of digits.

Given a number with N digits, the probability that it doesn't have any 9s is (9/10)^N because each digit is equally likely at a position.

(9/10)^N becomes very small as N becomes large, so the probability of a large number not having 9s is 0.


It's only 0 as far as statisticians and cheap calculators are concerned. The number of N-digit numbers without 9s in them is much higher than the number of similar (N-1)-digit numbers, it's just the proportion that falls. The probability isn't 0, it just asymptotically approaches 0, as the number of non-9-containing numbers approaches infinity.

In essence, the cardinality of the set of all numbers that do not contain 9 is infinity; the cardinality of the set of all numbers is infinitier. This despite the fact that both sets are countably infinite.


The way an infinitely unlikely but theoretically possible event is described is in fact 'probability 0'.

Sure, in a specific finite range the chance is a specific positive number, but the overall chance is 0.


Cardinality is not the term you want to be using.


I can see how it's imprecise, but I'm struggling to find a better descriptor. "Size" doesn't really work.


So, this is just a parlor trick, right? It sounds like this is just a complicated way of saying that this is what the harmonic series looks like when you've removed all terms with more than N digits.


Not really, because you've removed 'almost all' terms, but there are many ways to remove 'almost all' terms that do not result in a convergent series. For instance, you could remove 999 out of every 1000 terms. I'd say that counts as removing 'almost all' terms. Nevertheless, the resulting series does not converge, for the exact same reason that removing every other term doesn't work. The amount of terms that you remove has to increase faster than that.


And in particular, removing 100% of the terms doesn't mean much either, since, for example, the series 1/p for all primes does not converge, or 1/floor(n log n). You need to remove them harder than that.


well, almost all, yes.


How does that make terms with the 9 special?


It doesn't, this is true for any digit.


Which was my point ;-)


Can you elaborate on that? I can't find any point or contribution made by that post if it wasn't an actual question.


HSO's original comment was sarcasm posed as a question. Poorly executed, but that's where his original point was made ;)


Well yes, I understand that it was sarcasm. But while looking at it as sarcasm I don't see any point actually being made.


You can remove 8, 7 , 1 ,2... in fact, any digit and that will make the series convergent. You can go further. Remove 'any' group of digits that occurs in the series repeatedly (say, all terms with 373737 somewhere in it) and the series will converge. This is because as the terms get bigger, it becomes increasingly common to find your chosen 'digit' in almost all the terms, thus making the series convergent.


"as the terms get bigger, it becomes increasingly common to find your chosen 'digit' in almost all the terms, thus making the series convergent."

No! I think that condition is necessary, but it certainly is not sufficient for the series to converge. For example, non-primes become increasingly common, but the sum of their reciprocals is divergent.


I don't think gsk was claiming it was a sufficient condition -- just trying to give some intuition for why the phenomenon happens. (After all, if you only look at very small numbers is seems like most don't have 9s.)


There is a discussion of this over at /r/math: http://www.reddit.com/r/math/comments/if4kd/take_the_harmoni...




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