Rationals r(n):= A120794(n)/ A120795(n) = =sum(((-1)^k)*C(k)/16^k,k=0..n) with the Catalan numbers C(k)=A000108(k). r(n) for n=0..30: [1, 15/16, 121/128, 3867/4096, 30943/32768, 495067/524288, 3960569/4194304, 253475987/268435456, 2027808611/2147483648, 32444935345/34359738368, 259559486959/274877906944, 8305903553295/8796093022208, 66447228478363/70368744177664, 1063155655468083/1125899906842624, 8505245244078969/9007199254740992, 1088671391232413187/1152921504606846976, 8709371129876984331/9223372036854775808, 139349938077966926901/147573952589676412928, 1114799504623854824883/1180591620717411303424, 35673584147962470764661/37778931862957161709568, 285388673183701407147393/302231454903657293676544, 4566218770939216397791533/4835703278458516698824704, 36529750167513742617652719/38685626227668133590597632, 2337904010720879355999967191/2475880078570760549798248448, 18703232085767035170475774359/19807040628566084398385987584, 299251713372272561512125789381/316912650057057350374175801344, 2394013706978180494392925449067/2535301200456458802993406410752, 76608438623301775803190226641143/81129638414606681695789005144064, 612867508986414206458490307097939/649037107316853453566312041152512, 9805880143782627303210564636485603/10384593717069655257060992658440192, 78447041150261018425922953748265593/83076749736557242056487941267521536] This is the first member (p=1) of the fourth p-family of scaled Catalan sums. See the W. Lang link under A120996. The values of some partial sums r(n) of the convergent series sum(((-1)^k)*C(k)/16^k,k=0..infty) are (maple10 10 digits): [.9375000000, .9442719127, .9442719100, .9442719100] for n=10^k with k=0..3. This should be compared with the limit 4*(sqrt(5)-2) = 4*(2*phi-3)) = 4/phi^3 = 0.944271909... The sum sum(C(k)/(-16)^k,k=0..infinity) is convergent due to Leibniz' criterion because {C(k)/16^k} is a monotonely decreasing 0-sequence. The latter fact follows from the convergence of sum({C(k)/16^k,k=0..infinity), which can be shown with J. L. Raabe's criterion (cf. A120784(n)/ A120785(n) and the W. Lang link there). ################################################################################################################################ ######################################################################################################################################