= Characteristic functions of symmetric sign sums
{title2=$\cos(t/n)^n\to1,\quad\cos(t/\sqrt n)^n\to e^{-t^2/2}$}
Let $X_1,X_2,\ldots$ be <independent and identically distributed random variables> taking $-1$ and $1$ with equal <probabilities>, the <Rademacher distribution>. The two opposite <complex exponentials> give $\varphi_X(t)=(e^{it}+e^{-it})/2=\cos t$. Their <expected value> is zero and their <variance> is one. For $S_n=\sum_{j=1}^nX_j$, the <characteristic functions> of its <sample mean> and its <variance normalization> are respectively
$$
\varphi_{S_n/n}(t)=\cos(t/n)^n,\qquad
\varphi_{S_n/\sqrt n}(t)=\cos(t/\sqrt n)^n.
$$
For each fixed $t$, the bases are positive for all sufficiently large $n$. The <Taylor expansion> $\log\cos u=-u^2/2+O(u^4)$ at zero then gives
$$
n\log\cos(t/n)=-\frac{t^2}{2n}+O(n^{-3})\longrightarrow0,
\qquad
n\log\cos(t/\sqrt n)=-\frac{t^2}{2}+O(n^{-1})\longrightarrow-\frac{t^2}{2}.
$$
Exponentiating proves the two displayed pointwise limits. The first limit is the <characteristic function> of the constant zero; the second is that of the <standard normal distribution>. The <Lévy continuity theorem> gives the corresponding <convergence in distribution> statements. Averaging has <variance> $1/n$, whereas dividing the sum by $\sqrt n$ preserves <variance> one. This sign calculation illustrates why the <law of large numbers> and the <central limit theorem> use different scales; it is not a proof for arbitrary <probability distributions>.
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