Errata:

Change

P(X(i) = -1 = P(X(i) = 1 ) = 1/2

to

P(X(i) = -1 ) = P(X(i) = 1 ) = 1/2

For more, let v(i) be the variance of X(i). Then since E[X(i)] = 0

  v(i) = E[ (X(i) - E[X(i)])^2 ]

       =  E[ (X(i))^2 ]

       = (1/2) 1 + (1/2) 1

       = 1

Since in the i.i.d. case the variance of a sum is the sum of the variances, the variance of S(n) is n. Then the standard deviation of S(n) is sqrt(n) so that the standard deviation of

(1/sqrt(n)) S(n)

is 1.

So, for large n, the density of

(1/sqrt(n)) S(n)

converges to Gaussian with expectation 0 and standard deviation 1 and variance 1.