Yeah, quite like what Fred said: Take every number until the sum of the reciprocals exceeds 1, then every second number until the running sum of reciprocals exceeds 2, then every 3rd until it exceeds 3, etc. On Wed, Jun 26, 2013 at 9:21 PM, Fred W. Helenius <fredh@ix.netcom.com>wrote:
On 6/26/2013 8:23 PM, Keith F. Lynch wrote:
Last month, Zhang proved that there exists a number N such that there are infinitely many primes that differ from another prime by not more than N. (He showed that N is at most 70 million. That upper bound has since been reduced to 12,012. See http://michaelnielsen.org/**polymath1/index.php?title=** Bounded_gaps_between_primes<http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes> )
I've wondered if the same is true for any monotonically increasing sequence of positive integers (i.e. no duplicate terms) for which the sum of the reciprocals diverges. Can anyone think of a counterexample?
Here's a counterexample. Take the sequence that consists of all the 1-digit numbers, every other 2-digit number, every third 3-digit number, etc. There's some freedom in choosing the first entry of each length; do it so that the sequence of first differences is non-decreasing. For example:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, ..., 96, 98, 100, 103, ..., 994, 997, 1000, 1004, ..., 9992, 9996, 10000, 10005, ..., 99990, 99995, 100000, 100006, ..., 999988, 999994, 1000000, 1000007, ..., 9999991, 9999998, 10000005, 10000013, ...
(It took a while before 10^n wasn't in the arithmetic progression on the previous line!)
The first differences grow, and the sum of the reciprocals diverges because we have (roughly) 0.9*10^n / n terms on the nth line, each less than 10^n, so their reciprocals add up to more than 0.9/n.
-- Fred W. Helenius fredh@ix.netcom.com
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