CF = 2F1 identities typically have two contiguous 2F1s. These two were no exception! But Underpublicized Fact (apparently known by Mathematica): If a or b is a positive integer, 2F1[a,b;c,z] obeys a *two* term (inhomogeneous) recurrence vs the usual three term. Thus we can eliminate all but one species. On Mon, Sep 24, 2012 at 10:07 PM, Bill Gosper <billgosper@gmail.com> wrote:
An infinite continued fraction with fractional terms approaching 1. Out[123]= 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 {-, -, -, -, -, -, -, -, --, --, --, --, --, --, --, --, --, --, --, --, 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
69 ... , --} == 70
1 Sqrt[5] 1 1 1 -(-) + ------- + - Hypergeometric2F1[1, 1, 2 - -------, -- (5 - Sqrt[5])] 2 2 2 Sqrt[5] 10
Check: In[124]:= N[FromContinuedFraction[%[[1]]] -> %[[2]], 33]
Out[124]= 1.23225970948789169611305410061559 -> 1.23225970948789169611305410135170 Elementwise reciprocating the terms,
In[125]:= 1 ------------- == Out[95][[2]] Out[123][[1]]
Out[125]= 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 {2, -, -, -, -, -, -, -, --, --, --, --, --, --, --, --, --, --, --, --, 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
70 ..., --} == 69
(5 + 3 Sqrt[5]) /
1 (5 + Sqrt[5] - Sqrt[5] Hypergeometric2F1[1, 1, 2 + -------, Sqrt[5]
1 -- (5 - Sqrt[5])]) 10
In[126]:= N[FromContinuedFraction[%[[1]]] -> %[[2]], 33]
Out[126]= 2.49245974602128660339685183239050 -> 2.49245974602128660339685183239151 Larger terms -> more digits.
The infinitely repeated term sequence { 2i, 2i, 2i, ...} converges simply to i. The sequence with nth term 2i+1/n, as opposed to 1+1/n, {1 + 2 I, 1/2 + 2 I, 1/3 + 2 I, 1/4 + 2 I, 1/5 + 2 I, 1/6 + 2 I, 1/7 + 2 I, 1/8 + 2 I, 1/9 + 2 I, 1/10 + 2 I, 1/11 + 2 I, 1/12 + 2 I, 1/13 + 2 I, 1/14 + 2 I, 1/15 + 2 I, 1/16 + 2 I, 1/17 + 2 I, 1/18 + 2 I, 1/19 + 2 I, 1/20 + 2 I, 1/21 + 2 I, 1/22 + 2 I, 1/23 + 2 I, 1/24 + 2 I, 1/25 + 2 I, 1/26 + 2 I, 1/27 + 2 I, 1/28 + 2 I, 1/29 + 2 I, 1/30 + 2 I, 1/31 + 2 I, 1/32 + 2 I, 1/33 + 2 I} == 1/(-I + I E^-I (\[Pi] + I ExpIntegralEi[I])) This is weird because the rhs appears to be of lower computational complexity than the 2F1[(5-√5)/10] from the 1+1/n CF. The former is "T2" (confluent, e-like) while the latter appears to be 'T3", where Tk := n (log n)^k log log n, for n digits. (Note that all known π series are T3, but π is T2 via AGM, etc. Maybe this means there's a T2 algorithm for the 2F1[(5-√5)/10].) The CF {i,i,i,...} is meaningless because In[40]:= FromContinuedFraction[{I, I, I, a}] Out[40]= a I.e., three consecutive "i"s are like not there. But what about the CF which merely approaches {i,i,i,...}? Formally {1 + I, 1/2 + I, 1/3 + I,...} = 1/cfk[1, I + 1/n, {n, 1, ∞}] == (-1)^(5/6)/( 1 - Hypergeometric2F1[1, 1, 1 + 1/Sqrt[3], 1/6 (3 - I Sqrt[3])]), but does this actually converge? An all-night run (1.1 Gterms) confirmed the identity with an error of 2.25/10^10. --rwg