I have an algebra question.
I am currently working on this function:
$$\left(\frac{{e^{t\sqrt{3/n}}} -{e^{-t\sqrt{3/n}}}}{2t\sqrt{3/n}}\right)^n$$
Now this is the MGF of a function called $Z$ and I have to find the MGF as $n$ goes to $\infty$ which of course, since I am using the CLT, will have to be $e^{t^2/2}$ (the MGF of the Standard Normal). All of this is correct, it's just that I am having difficulty simplifying this. I also know that I need to use the following:
$(1+a/n)^n$ as $n$ goes to $\infty$ is equal to $e^a$.
Help would be greatly appreciated as I am quite unsure of how to proceed. I do not even know where to begin as the function is raised to the power of $n$.
Thanks so much!
With $y=t \sqrt{\frac{3}{n}}\;$ your expression becomes $$f(t,n) := \left(\frac{e^y - e^{-y}}{2y}\right)^n = \left(\frac{\sinh y}{y}\right)^n$$ Now for large $n$ i.e. small $y$ we use the Taylor series $$\frac{\sinh y}{y} = 1+\frac{1}{6}y^2 + \frac{1}{120}y^4+O(y^5)$$ It is easier to approximate $\ln f(t,n)\;$ using an other Taylor series (for small $x$): $$\ln(1+x)= x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+O(x^5)$$ With these definitions and tools we can get $$\ln f(t,n) = \ln \left(\left(\frac{\sinh y}{y}\right)^n\right) =n \ln \left(\frac{\sinh y}{y}\right) \approx n \ln \left(1+\frac{1}{6}y^2 + \frac{1}{120}y^4+O(y^5)\right) \approx n \left(\frac{1}{6}y^2 + O(y^3)\right) =n \left(\frac{1}{6}t^2\frac{3}{n} + O(n^{-3/2})\right) =\frac{1}{2}t^2 + O(n^{-1/2}) $$ and this will give the approximation $e^{\frac{1}{2}t^2}$.
Please note that this is the basic idea; with more terms in the series you will get better error estimates.