The problem states: Let $(X_{j})_{j>1}$ be i.i.d. with $P(X_{j}=1) = P(X_{j}=0)=\frac{1}{2}$. Let $S_{n}=\sum_{j=1}^{n}X_{j}$, and let $Z_{n}=2S_{n}-n$.($Z_{n}$ is in excess of heads over tails in $n$ tosses, if $X_{j}=1$ when heads on the $j$th toss, and $X_{j}=0$ when tails on the $j$th toss.) Show that $\lim_{n \to \infty} P\left(\frac{Z_{n}}{\sqrt{n}}<x\right)= \Phi (x)$, where $\Phi (x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}\exp(-u^{2}/2)du$.
This is what I did:
$P\left(\frac{Z_{n}}{\sqrt{n}}<x\right)=P\left(\frac{2S_{n}-n}{\sqrt{n}}<x\right)$. By the Central Limit Theorem, we have that $\frac{2S_{n}-m}{\sqrt{n}} \to ^{\mathcal{D}} N(0,1)$. So, $P\left( \frac{1}{\sqrt{n}}(2S_{n}-n)<x \right) \to^{\mathcal{D}}\int_{-\infty}^{x}\exp^{-u^{2}/2}du=\Phi(x)$.
Is this all that needs to be done for this problem, or is there more that I need to show? If so, what specifically do I need to show? It just seems too easy...