The problem is as follows: Given two players, each flips a coin that produces heads with probability $p$. This is repeated $n$ times. What is the asymptotic behavior of the expected absolute total difference of heads?
This problem has numerous, slightly less general variants that have been "solved" online using the central limit theorem. Unfortunately, convergence in distribution does not guarantee convergence of means, yet this seems to be a common assumption in using normal distributions to approximate such a mean. The argument is as follows:
Suppose $\{X_i\}$ and $\{Y_i\}$ are both sets of $n$ Bernoulli random variables, then by the central limit theorem,
\begin{align} \frac{\sum_{i=1}^n X_i-np}{\sqrt{np(1-p)}} \overset{d}{\longrightarrow} \mathcal{N}(0,1) \end{align} Taking the difference of normals and then folding the distribution gives \begin{align} \left| \frac{\sum_{i=1}^n X_i - \sum_{i=1}^n Y_i}{\sqrt{np(1-p)}} \right| \overset{d}{\longrightarrow} \mathcal{N}\left(\frac{2}{\sqrt{\pi}},2\left(1-\frac{2}{\pi}\right)\right) \end{align}
I would like to conclude that \begin{align} \lim_{n\to\infty} \frac{\mathbb{E}\left|\sum_{i=1}^n X_i - \sum_{i=1}^n Y_i\right|}{\sqrt{n}} = 2\sqrt{\frac{p(1-p)}{\pi}} \end{align}
Attempt
To strengthen convergence in distribution to convergence of means it suffices to show uniform integrability. I found an explicit formula for the expected value at each $n$ using a random walk and the law of total expectation.
\begin{align} \mathbb{E}\left| \sum_{i=1}^n X_i - \sum_{i=1}^n Y_i\right| = 2p(1-p) \sum_{i=0}^{n-1} \sum_{j=0}^i \binom{i}{j}^2 p^{2j}(1-p)^{2i-2j} \end{align}
In the special case of $p=\frac{1}{2}$ this has a simple form: \begin{align} \frac{1}{\sqrt{n}}\mathbb{E}\left| \sum_{i=1}^n X_i - \sum_{i=1}^n Y_i\right| = \frac{1}{2\sqrt{n}} \sum_{i=0}^{n-1}\frac{1}{2^{2i}}\sum_{j=0}^i \binom{i}{j}^2 = \frac{1}{2\sqrt{n}} \sum_{i=0}^{n-1} \frac{\binom{2i}{i}}{4^i} = \frac{\Gamma(n+\tfrac{1}{2})}{\sqrt{\pi n} \Gamma(n)} \to \frac{1}{\sqrt{\pi}} \end{align}
I could not find a simple closed form for general $p$ however. Is there an easier way to show uniform integrability, so that the CLT result may be justified?
Edit
The explanation of the calculated mean is as follows:
If we suppose that $X_i$ and $Y_i$ represent whether a step is taken in the (0,1) or (1,0) direction on a lattice grid. The excess $|\sum_{i=1}(X_i-Y_i)|$ may then be seen to be $|a-b|$, where the path ended on $(a,b)$.
For any path that ends off the diagonal (i.e., when we end on $(j,k)$ for $j\neq k \in \{0,1,\dotsc,n\}$), the mean increment is $0$. This is true because with probability $p^2+(1-p)^2$ the excess is zero, with probability $p(1-p)$ the excess decreases by $1$, and with probability $p(1-p)$ the excess increases by $1$.
For any path that ends on the diagonal (i.e., when we end on $(k,k)$ for some $k \in \{0,1,\dotsc,n\}$), then with probability $p^2+(1-p)^2$ the excess is zero, and with probability $2p(1-p)$ the excess increases by $1$, i.e., $(k+1,k)$ and $(k,k+1)$ are both increase excess by 1.
The summation then comes from counting all walks terminating on the diagonal before the $n$th movement, where the mean increments are non-zero.
Final Solution
Let $\varepsilon_n = \sum_{i=1}^n X_i - \sum_{i=1}^n Y_i$. Let $G(t)=t^2$, then by de la Vallée-Poussin theorem, $\left\{\frac{\varepsilon_n}{\sqrt{n}}\right\}$ is uniformly integrable if
\begin{align} \sup_n \mathbb{E}\left[ \left|\frac{\varepsilon_n}{\sqrt{n}}\right|^2 \right] < \infty \end{align}
We calculate $\mathbb{E}[\varepsilon_n^2]$ explicitly. Noting that $\varepsilon_n = \varepsilon_{n-1}+X_{n-1}-Y_{n-1}$,
\begin{align} \mathbb{E}[\varepsilon_n^2] = \mathbb{E}[\varepsilon_{n-1}^2] + 2\mathbb{E}[\varepsilon_{n-1}] \mathbb{E}[X_{n-1}-Y_{n-1}] + \mathbb{E}[ (X_{n-1}-Y_{n-1})^2 ] \end{align}
Now, because $X_{n-1}$ and $Y_{n-1}$ are i.i.d, it follows that $\mathbb{E}[X_{n-1}-Y_{n-1}]=0$.
We find that $(X_{n-1}-Y_{n-1})^2 = 1$ if $X_{n-1}=1$ and $Y_{n-1}=0$, or if $X_{n-1}=0$ and $Y_{n-1}=1$, each of which occurs with probability $p(1-p)$, hence \begin{align} \mathbb{E}[\varepsilon_n^2] = \mathbb{E}[\varepsilon_{n-1}^2] + 2p(1-p) = 2np(1-p) \end{align}
It follows that \begin{align} \mathbb{E}\left[ \left|\frac{\varepsilon_n}{\sqrt{n}}\right|^2 \right] = \frac{1}{n}[2np(1-p)] = 2p(1-p) < \infty \end{align}
Uniform integrability strengthens convergence in distribution to convergence in mean, which implies convergence of means.