Nikhil, your method of evaluating the two infinite series will not work for the sizes of the values in 'n'. Factorial(2e4) is an indescribably large value far, far outside the range of matlab's double precision numbers. You should be evaluating the quantities in your series iteratively in terms of their previous values in 'n'. For example the term in the series for F with n = 3 is:
((1*3*5)/(2*4*6))^2*k^6
To get to the next term with n = 4, the above can be multiplied by
(7/8)^2*k^2
to arrive at
((1*3*5*7)/(2*4*6*8))^2*k^8
Generalizing this technique you can avoid the horrendous sizes of the factorials when n becomes large.
(Note: I am assuming you meant {[(2n)!/(2^(2n)*(n!)^2)]^2}*[k^(2n)] in your expression, as in your previous posting.)
