

The harmonic series, after removing any term with a "9" in it, is convergent. - japaget
http://hippomath.blogspot.com/2011/07/kempner-series-modified-harmonic-series.html

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Almaviva
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.

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endtime
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.

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

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

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radq
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.

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redthrowaway
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.

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neutronicus
Cardinality is not the term you want to be using.

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

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gsk
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.

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Someone
_"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.

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gjm11
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.)

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radq
There is a discussion of this over at /r/math:
[http://www.reddit.com/r/math/comments/if4kd/take_the_harmoni...](http://www.reddit.com/r/math/comments/if4kd/take_the_harmonic_series_remove_any_term_with_a_9/)

