Let Hn be the number of heads when flipping n fair coins let

Let H_n be the number of heads when flipping n fair coins, let X_n=e^(-H_n) , and let Y=0. Prove that X_n ----> Y

Solution

we have that Hn is the number of heads whne plipping n fair coins

as we know that number will always n/2 because is fair

so if we use Xn as e^- Hn

we are supposing that Xn will be e^- n/2

but if n increases a lot of we will suppose that the number of Hn will increase so

e^- n when n is big

will be closer than 0

and Y = 0

so X_n = Y