Geometric Brownian Motion

Hi, I am trying to answer the following question:



Consider a geometric Brownian motion S(t) with S(0) = S_0 and parameters μ and σ^2. Write down an approximation of S(t) in terms of a product of random variables. By taking the limit of the expectation of these compute the expectation of S(t).



Let Z_i = σ√Δ (with probability 1/2(1+ [itex]\frac{μ}{σ}[/itex]√Δ)) or -σ√Δ (with probability 1/2(1- [itex]\frac{μ}{σ}[/itex]√Δ)).



Then logS(t) ≈ Z_1 + ... + Z_([itex]\frac{t}{Δ}[/itex]) (as logS(t) is Brownian motion)



And this approximation gets better as [itex]\frac{t}{Δ}[/itex] tends to infinity



S(t) ≈ exp(Z_1) * ... * exp(Z_([itex]\frac{t}{Δ}[/itex]))



So now exp(Z_i) = exp(σ√Δ) (with probability 1/2(1+ [itex]\frac{μ}{σ}[/itex]√Δ)) or exp(-σ√Δ) (with probability 1/2(1- [itex]\frac{μ}{σ}[/itex]√Δ)).



E[S(t)] ≈ (E[exp(Z_1)])^[itex]\frac{t}{Δ}[/itex]



I know that the answer I am looking for is S_0*exp(tμ+[itex]\frac{tσ^2}{2}[/itex]). Also



E[exp(Z_1)] =1/2exp(σ√Δ)*(1+ [itex]\frac{μ}{σ}[/itex]√Δ) -1/2exp(-σ√Δ)*(1- [itex]\frac{μ}{σ}[/itex]√Δ).



So I believe that



E[S(t)] = lim [ 1/2exp(σ√Δ)*(1+ [itex]\frac{μ}{σ}[/itex]√Δ) -1/2exp(-σ√Δ)*(1- [itex]\frac{μ}{σ}[/itex]√Δ) ]^[itex]\frac{t}{Δ}[/itex] (as [itex]\frac{t}{Δ}[/itex] tends to infinity)



Where do I go from here? I want to make use of lim (1+x/t)^t = exp(x) but how do I go from here? Help is MUCH appreciated!