I know the central limit theorem can be proved very easily by looking at characteristic functions. But I'm finding it very difficult to prove directly for the case of a sequence of independent coin tosses.
Precisely, if $(\Omega, P)$ is a probability space, $X_i:\Omega \to \{\pm 1\}$ are independent with $P[X_i = 1] = 0.5$ (coin tosses), and $S_n = X_1 + \dots + X_n$, I want to show that the measures on $\mathbb{R}$ induced by $S_n/\sqrt{n}$ converge to the standard normal. That is, for fixed $a < b$, I want to show $$\lim\limits_{n\to \infty} \sum\limits_{a \leq \frac{2k-n}{\sqrt n} \leq b} \frac{1}{2^n}{n \choose k} = \int\limits_{a\leq x \leq b}\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{x^2}{2}\right]dx. $$ Note, $S_n$ can take the values $2k-n$ with $k=0,\dots,n$ which is where the conditions on the sum are coming from.
The attempt is to use Stirling's formula and somehow write the left side as a Riemann integral. I'm not sure how to make this work however, since the division of the interval $[a,b]$ in the sum is by points $\frac{2}{\sqrt{n}}$ apart and I need help rewriting the summands in this fashion.
Fix $-\infty < a < b < \infty$ and define $t_k = t^{(n)}_k$ by the relation $k = \frac{n}{2} + \frac{\sqrt{n}}{2} t_k $ for $k \in \{0,\dots,n\}$. Then by the Stirling's approximation,
\begin{align*} \binom{n}{k} &= \frac{1 + \mathcal{O}(n^{-1})}{\sqrt{2\pi n}} \left( \frac{k}{n} \right)^{-(k+\frac{1}{2})}\left( \frac{n-k}{n} \right)^{-(n-k+\frac{1}{2})} \\ &= (1 + \mathcal{O}(n^{-1})) \frac{2^{n+1}}{\sqrt{2\pi n}} \left( 1 + \frac{t_k}{\sqrt{n}} \right)^{-\frac{n+t_k\sqrt{n}+1}{2}} \left( 1 - \frac{t_k}{\sqrt{n}} \right)^{-\frac{n-t_k\sqrt{n}+1}{2}} \\ &= (1 + \mathcal{O}(n^{-1/2})) \frac{2^{n+1}}{\sqrt{2\pi n}} e^{-\frac{1}{2}t_k^2} \end{align*}
uniformly in $k$ satisfying $a \leq t_k \leq b$. Then by noting that $t_k - t_{k-1} = \frac{2}{\sqrt{n}}$,
$$ \sum_{k : a \leq t_k \leq b} \binom{n}{k}\frac{1}{2^n} = \frac{1 + \mathcal{O}(n^{-1/2})}{\sqrt{2\pi}} \sum_{k : a \leq t_k \leq b} e^{-\frac{1}{2}t_k^2} (t_k - t_{k-1}), $$
and letting $n\to\infty$ proves the desired equality.
Remark. The mode of convergence mentioned in the above proof is not the same as the converge in distribution, at least on the face of it. In fact, this corresponds to something called vague convergence. However, we can prove that the above statement is sufficient to establish the convergence in distribution of $S_n/\sqrt{n}$ to the standard normal variable, by taking advantage of the fact that all the measures involved are probability measures.