Rationals r(n):= A120788(n)/ A120777(n) = sum(C(k)/(-4)^k,k=0..n).
 
 r(n):=sum(((-1)^k)*C(k)/4^k,k=0..n) with C(k):=A000108(k) (Catalan numbers).
 
 r(n) for n=0..30:

 [1, 3/4, 7/8, 51/64, 109/128, 415/512, 863/1024, 13379/16384, 27473/32768, 107461/131072, 219121/262144, 1723575/2097152, 
  3499153/4194304, 13810887/16777216, 27956079/33554432, 884899683/1073741824, 1787478201/2147483648, 7085090409/8589934592, 
  14289590493/17179869184, 113433092349/137438953472, 228507214803/274877906944, 907912292457/1099511627776, 
  1827259905369/2199023255552, 29064628679079/35184372088832, 58451733394989/70368744177664, 232591446979593/281474976710656, 
  467478813093205/562949953421312, 3722447117016639/4503599627370496, 7477862728002073/9007199254740992, 
  29786170634926871/36028797018963968, 59810777926234511/72057594037927936]

 The sum sum(C(k)/(-4)^k,k=0..infinity) is convergent due to Leibniz' criterion because {C(k)/4^k} is a monotonely decreasing 
 0-sequence. The latter fact follows from the convergence of sum({C(k)/4^k,k=0..infinity), which can be shown with 
 J. L. Raabe's criterion (cf. A120778(n)/ A120779(n) and the W. Lang link there).
  
 
 The values of some partial sums r(n) of the convergent series sum(C(k)/(-4)^k,k=0..infty) are (maple10 10 digits):

 [.7500000000, .8358802795, .8287040012, .8284360287] , k=0,...,3.

 
 This should be compared with the limit 2*(sqrt(2)-1) = 0.828427124...


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