Re: Mysterious series

David Bernier wrote:
Gerard Schildberger wrote:
[...]

How many terms do you have Mathematica sum? I used my own calculator
and for

88 terms, I got 1.5166
100 terms, I got 1.4883
200 terms, I got 1.5173

each calculated with 388 digits of accuracy. __________________Gerard S.

With the method of starting with 1/pi (50,000 digits of
1 million, from the Web, sourced to Kanada in Japan), I got:

k term sum to k'th term
--------------------------------------------
88 0.007228908018 1.516642155847
100 -0.001716064335 1.488312766235
200 -0.000856044241 1.517313876254
300 0.000434104752 1.486303622743
400 0.000477834061 1.489514348740
500 -0.001787464191 1.516076015695
600 -0.001096000006 1.489092411522
700 -0.001372388315 1.490684934440
800 -0.001046091143 1.497726495519
900 -0.001058517094 1.490134592658
1000 -0.000972800356 1.491758126951
2000 -0.000488800614 1.481082589871
3000 -0.000295414348 1.487370295823
4000 -0.000228178495 1.484784976933
5000 -0.000198556946 1.483166697635
6000 0.000021272601 1.476767865976
7000 -0.000086441949 1.475627052336
8000 0.000123621407 1.472164450790
9000 -0.000064422362 1.466283910081
10000 -0.000098994212 1.466036462288

I don't have enough digits to go to 20,000 or beyond...

junk
||
\/
20000 0.000030475692 1.476894852357
30000 -0.000021766162 1.476968078448
40000 0.000024901165 1.472950395550


P.S.: I use the fact that
frac( (n+1)!/2pi ) = frac((n+1) frac(n!/2pi)).

David Bernier

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