Binomial distribution $n \rightarrow \infty$

Question

Which is the value of

$$ \lim_{n\rightarrow \infty} \sum_{x=0}^{x=n/2} \varepsilon^{2x}(1-\varepsilon)^{n-2x} \frac{n!}{(2x)!(n-(2x))!} $$

i.e the limit of the Binomial distribution summed over all even values of an also even $n$ ($0, 2, 4,\ldots$)? I suppose that it should give $1/2$. Can someone tell me if it is true?

Answer 1

First, taking the even part of $(1+x)^n$ using the Binomial Theorem: $$ \begin{align} \sum_k\binom{n}{2k}x^{2k} &=\frac12\left[\sum_k\binom{n}{k}x^k+\sum_k\binom{n}{k}(-x)^k\right]\\[6pt] &=\frac12\left[(1+x)^n+(1-x)^n\right] \end{align} $$ Therefore, $$ \begin{align} \sum_{k=0}^n\binom{n}{2k}\varepsilon^{2k}(1-\varepsilon)^{n-2k} &=(1-\varepsilon)^n\sum_{k=0}^n\binom{n}{2k}\left(\frac\varepsilon{1-\varepsilon}\right)^{2k}\\ &=\frac12(1-\varepsilon)^n\left[\left(1+\frac\varepsilon{1-\varepsilon}\right)^n+\left(1-\frac\varepsilon{1-\varepsilon}\right)^n\right]\\[3pt] &=\frac12\left[1+(1-2\varepsilon)^n\right] \end{align} $$ So, for $0\lt\varepsilon\lt1$, we get the limit of $\frac12$. Note that this excludes $\varepsilon=0$ and $\varepsilon=1$.