This applies to any divergent series, right? And the fun thing is that any C has a series with any initial subsequence. I suppose this means any C has countably many sequences, although it’s not clear that taking a finite initial subsequence and then turning greedy covers all of them… - Cris
On Dec 10, 2018, at 3:43 PM, Dan Asimov <dasimov@earthlink.net> wrote:
There is the greedy-harmonic-series-with-signs method:
Let C be the irrational real number to approximate; say C > 0.
Then let p1 be the least integer for which
1 + 1/2 + 1/3 + ... + 1/p1 > C.
Now let n1 be the least integer > p1 for which
1 + 1/2 + 1/3 + ... + 1/p1 - (1/(p1+1) + 1/(p1+2) + ... + 1/n1) < C
Continue this way, alternately adding and subtracting groups of consecutive terms, resulting in a signed harmonic series that converges to C.
—Dan
Keith Lynch wrote: ----- ..... What are some exotic ways to approximate real numbers to any desired precision? Extra points if you can come up with something completely original. -----
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