# Asymptotic Normality of the Median

I'm asking this question based off of Example 10.2.3 in Casella & Berger. We have a sample $X_1, ..., X_n$ from a population with pdf $f$ and (differentiable) cdf $F$. Let $\mu$ be the population median and $M_n$ be the sample median. We wish to compute

$$\lim_{n \to \infty} P(\sqrt{n}(M_n - \mu) \leq a)$$

for some $a$. We then define $Y_i$ by

$$Y_i = \begin{cases} 1 & \mbox{if } X_i \leq \mu + a/\sqrt{n} \\ 0 & o.w. \end{cases}$$

Clearly the $Y_i$ are Bernoulli with success probability $p_n = F(\mu + a/\sqrt{n})$. If we assume that $n$ is odd, then $\{M_n \leq \mu + a/\sqrt{n}\}$ is equivalent to $\{\sum Y_i \geq (n+1)/2\}$. Now, the book states "A straightforward limit calculation will also show that"

$$\frac{(n+1)/2 - np_n}{\sqrt{np_n(1-p_n)}} \to -2aF'(\mu) = -2af(\mu)$$.

I follow completely up until this limit calculation at which point I'm completely lost. I tried substituting in the definition of $p_n$, separating, etc. I'm probably just overlooking something simple but this "straightforward" calculation has been eluding me for a couple hours now.