r(n):= A120791(n)/A120796(n) = sum(((-1)^k)*C(k)/20^k,k=0..n) with C(k):=A000108(k) (Catalan numbers). r(n), n=0..30: [1, 19/20, 191/200, 1527/1600, 76357/80000, 1527119/1600000, 15271223/16000000, 1221697411/1280000000, 488678993/512000000, 244339494069/256000000000, 2443394944889/2560000000000, 97735797766167/102400000000000, 977357977713673/1024000000000000, 3909431910817547/4096000000000000, 39094319108242331/40960000000000000, 6255091057316833991/6553600000000000000, 62550910573171875677/65536000000000000000, 1251018211463424549061/1310720000000000000000, 2502036422926853874509/2621440000000000000000, 500407284585370598175481/524288000000000000000000, 5004072845853706309960831/5242880000000000000000000, 100081456917074124975903269/104857600000000000000000000, 1000814569170741252046096781/1048576000000000000000000000, 16013033106731860025876356223/16777216000000000000000000000, 4003258276682965006791565092581/4194304000000000000000000000000, 80065165533659300134615815251257/83886080000000000000000000000000, 800651655336593001348454071646589/838860800000000000000000000000000, 32026066213463720053920779478134559/33554432000000000000000000000000000, 64052132426927440107848152655062877/67108864000000000000000000000000000, 6405213242692744010784689985229206279/6710886400000000000000000000000000000, 64052132426927440107847138288948443559/67108864000000000000000000000000000000] The values of some partial sums r(n) of the convergent series sum(((-1)^k)*C(k)/20^k,k=0..infty) are (maple10 10 digits): [.9500000000, .9544511503, .9544511501, .9544511501] for n=10^k with k=0..3. This should be compared with the limit (2*(sqrt(30)-5)) = 0.954451150.... The sum sum(C(k)/(-12)^k,k=0..infinity) is convergent due to Leibniz' criterion because {C(k)/12^k} is a monotonely decreasing 0-sequence. The latter fact follows from the convergence of sum({C(k)/12^k,k=0..infinity), which can be shown with J. L. Raabe's criterion (cf. A120786(n)/ A120787(n) and the W. Lang link there). ##########################################################################################################################