Rationals r(n) = A120996(n)/A120997(n), n>=0. r(n):= rI(p=1,n) = sum(C(k)/L(2)^(2*k),k=0..n), n>=0, with the Lucas number L(2)=3 and the Catalan numbers C(k):=A000108(k). r(n), n=0..30: [1, 10/9, 92/81, 833/729, 7511/6561, 22547/19683, 202967/177147, 1826846/1594323, 49326272/43046721, 443941310/387420489, 3995488586/3486784401, 35959456060/31381059609, 323635312552/282429536481, 2912718555868/2541865828329, 2912718853028/2541865828329, 26214470754457/22876792454961, 235930240718743/205891132094649, 6370116542620991/5559060566555523, 57331049042801819/50031545098999707, 515979441974304101/450283905890997363, 4643814979956777049/4052555153018976267, 41794334827766415781/36472996377170786403, 376149013480391929909/328256967394537077627, 1128447040479293524577/984770902183611232881, 10156023364456964404229/8862938119652501095929, 91404210280652895904889/79766443076872509863361, 7403741032751251921368161/6461081889226673298932241, 66633669294830800843229453/58149737003040059690390169, 599703023653740955540815437/523347633027360537213511521, 5397327212884670842083990301/4710128697246244834921603689, 48575944915965852565258005013/42391158275216203514294433201] ###################################################################################################################### The numerators A120996(n), n=0..30, are: [1, 10, 92, 833, 7511, 22547, 202967, 1826846, 49326272, 443941310, 3995488586, 35959456060, 323635312552, 2912718555868, 2912718853028, 26214470754457, 235930240718743, 6370116542620991, 57331049042801819, 515979441974304101, 4643814979956777049, 41794334827766415781, 376149013480391929909, 1128447040479293524577, 10156023364456964404229, 91404210280652895904889, 7403741032751251921368161, 66633669294830800843229453, 599703023653740955540815437, 5397327212884670842083990301, 48575944915965852565258005013] The denominators A120997(n), n=0..30, are: [1, 9, 81, 729, 6561, 19683, 177147, 1594323, 43046721, 387420489, 3486784401, 31381059609, 282429536481, 2541865828329, 2541865828329, 22876792454961, 205891132094649, 5559060566555523, 50031545098999707, 450283905890997363, 4052555153018976267, 36472996377170786403, 328256967394537077627, 984770902183611232881, 8862938119652501095929, 79766443076872509863361, 6461081889226673298932241, 58149737003040059690390169, 523347633027360537213511521, 4710128697246244834921603689, 42391158275216203514294433201] ####################################################################################################################### r(n) for n=10^k, k=0,1,2,3,4: (maple10, 10 digits) [1.111111111, 1.145894935, 1.145898034, 1.145898034, 1.145898034] This should be compared with the exact result: limit_{n->infinity}(r(n)) = 3*(2-phi) = 3/phi^2 = 1.145898034 (maple10, 10 digits) ######################################################################################################################### This is the p=1 member of the family r1(p,n), n>=0, p>=0, which is defined for general p as the partial sums of the normalized scaled Catalan series Csn1(p):=sum( C(k)/L(2*p)^(2*k),k=0..infinity). Among the general scaled Catalan series CS(q):=sum(C(k)/q^k,k=0..infinity) with natural q=4,5,6,.., with limit (q - sqrt(q*(q-4)))/2, this p-family Csn1(p) is one of two families (the other is called Csn2(p)) which have as limits integers in the quadratic number field Q(sqrt(5)). The coefficients turn out to be related to Fibonacci and Lucas numbers. If one uses unnormalized (i.e. k=0 term not 1) scaled Catalan sums, the relevant formulae for these two p-families are, with phi:=(1+sqrt(5))/2, the golden section and unit in Q(sqrt(5)): p-family I: sum(C(k)/L(2*p)^(2*k+1),k=0..infinity) = F(2*p+1) - F(2*p)*phi = 1/phi^(2*p) , p>=0, p-family II: sum(C(k)/((5^k)*F(2*p+1)^(2*k+1)),k=0..infinity) = L(2*p+2) - L(2*p+1)*phi = (2*phi-1)/phi^(2*p+1) , p>=0. For (unnormalized) alternating scaled Catalan sums one has the following two p-families: p-family III: sum(((-1)^k)*C(k)/((5^k)*F(2*p)^(2*k+1)),k=0..infinity) = -L(2*p+1) + L(2*p)*phi = (2*phi-1)/phi^(2*p), p>=1, p-family IV: sum(((-1)^k)*C(k)/L(2*p+1)^(2*k+1),k=0..infinity) = -F(2*p+2) + F(2*p+1)*phi = 1/phi^(2*p+1), p>=1. Here F(n):=A000045(n) (Fibonacci) and L(n):-A000032(m) (Lucas). Note that 2*phi-1 = sqrt(5). The families II and III give primes in Q(sqrt(5)) and have norm N= +5 and -5, respectively. The members of family II are also equal to (3-phi)*(1/phi)^(2*p), using phi^2=phi+1 or 1/phi=phi-1. Therefore,they give the squared lengths in the iterated golden triangle. 3-phi is the square of the length of the pentagon side if it is inscribed in a circle of radius 1. The first members of family II are: 3-phi, 7-4*phi, 18-11phi, 47-29*phi, 123-76*phi,... for p=0,...,4. The families I and IV (with p>=0) provide the positive units in Q(sqrt(5)) with norm N=+1 and -1, respectively. Beware that the family IV extrapolation to p=0 belongs to the divergent series sum(((-1)^k)*C(k),k=0..infinity). ######################################################################################################################## These results follow from the Taylor expansion sqrt(1+x) = 1 + (x/2)*sum(C(k)*(-x/4)^k),k=0..infnity) for |x|<=1. For the boundary value x=-1 see the remarks under A119951 and the W. Lang link there. For the boundary value x=+1 see remarks under A120788 and the W. Lang link there. a) For the scaled Catalan series CS(q) with positive terms one takes x=-4/q with q=4,5,6,... In order to have integers in the quadratic number field Q(sqrt(5)) as limits (q - sqrt(q*(q-4)))/2 one has to solve the Pell equation x^2 - 5*y^2 = +4 which has general (nonnegative) solutions (L(2*n),F(2*n)), n>=0. Then q=q(n)=2 + L(2*n), n>=0, and the limits become (F(2*n+1)+1) - F(2*n)*phi, with phi:=(1+sqrt(5))/2. Here one uses L(2*n)+F(2*n)=2*F(2*n+1) and sqrt(5)= 2*phi-1. For n even (n=2*p) and odd (n+2*p+1) one then obtains the two p-families I and II mentioned above. For this one used L(4*p) + 2 = L(2*p)^2 and L(4*p+2) + 2 = 5*F(2*p+1)^2. Also (F(2*n+1)+1) - F(2*n)*phi = L(n)*(F(n+1) - F(n)*phi) for even n, and = F(n)*(L(n+1) - L(n)*phi) for odd n were used. b) For the scaled Catalan series CS(q) with alternating terms one takes x=4/q with q=4,5,6,... In order to have integers in the quadratic number field Q(sqrt(5)) as limits (q - sqrt(q*(q-4)))/2 one has to solve also the Pell equation x^2 - 5*y^2 = +4 which has general (nonnegative) solutions (L(2*n),F(2*n)), n>=0. Then q=q(n)= -2 + L(2*n), n>=2, and the limits become -(F(2*n+)-1) + F(2*n)*phi, with phi:=(1+sqrt(5))/2. Here one uses also L(2*n)+F(2*n)=2*F(2*n+1) and sqrt(5)= 2*phi-1. For n even (n=2*p) and odd (n+2*p+1) one then obtains the two above mentioned p-families III and IV. For this one used L(4*p) - 2 = 5*F(2*p)^2 and L(4*p+2) - 2 = L(2*p+1)^2. Also -(F(2*n+)-1) + F(2*n)*phi = F(n)*(-L(n+1) + L(n)*phi) for even n, and = L(n)*(-F(n+1) + F(n)*phi) for odd n were used. ############################################## e.o.f. ###################################################################