CatalanNumber[z] \[Proportional] (2^(2 z)/(Sqrt[Pi] z^(3/2))) (1 - 9/(8 z) + 145/(128 z^2) - 1155/(1024 z^3) + 36939/(32768 z^4) - 295911/(262144 z^5) + 4735445/(4194304 z^6) - 37844235/(33554432 z^7) + 2421696563/(2147483648 z^8) - 19402289907/(17179869184 z^9) + O[1/z^10]) /; Abs[Arg[z + 1/2]] < Pi && (Abs[z] -> Infinity)

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2007-05-02