probability of winning a game

[Joshua Zucker]

To win a game, you must flip n+1 heads in a row, where n is the total
number of tails flipped so far.

Then the probability of winning for the first time after n tails is
A005329 / A006125 .

Can someone help me phrase this comment more clearly, or do you think
those two lines would be sufficient?  I could add an example: to win
when n = 2, you could have flipped TTHHH or THTHHH, hence the
probability is 3/64.  Or I could add more explanation: to avoid
winning at the earlier stages, you have a product of (2^k - 1)/2^k,
and then on the last stage you have a 1/2^(n+1) chance of winning, so
for example when n = 2 you have a 1/2 * 3/4 * 1/8 = 3/64 chance of
winning, and when n = 3 you have a 1/2 * 3/4 * 7/8 * 1/16 chance of
winning.  Hm, that part doesn't seem too clearly stated, but the
question is whether further explanation like that would be useful.

Also, is this of enough interest that I should submit the numerators
of the partial sums of this sequence (that is, the probability of
winning by flipping n+1 or fewer heads, somewhere along the way)?

 1,  5, 43, 709, 23003, 1481957, 190305691, 48796386661, 25003673060507, ...

And, finally, the limit of those partial sums (that is, the
probability of eventually winning) appears to be about
0.711211904913397578721100278...  can anyone help me figure out where
this number comes from?  (sum of A005329 / A006125, obtained by
brute-force-summing the first 150 terms; I'd rather see some pretty
math here!).

Thanks,
--Joshua Zucker