= Solution
For the <central limit theorem>, let $X_1,X_2,\ldots$ be <independent and identically distributed random variables> with <expectation> $\mu$ and finite positive <variance> $\sigma^2$. Then
$$
\boxed{\frac{X_1+\cdots+X_n-n\mu}{\sigma\sqrt n}\xrightarrow{d}N(0,1).}
$$
Thus the centered, normalized sum has <convergence in distribution> to the standard <normal distribution>. The positive-<variance> assumption is needed for this normalization; if $\sigma^2=0$, each summand equals $\mu$ <almost surely> and the centered sum is identically zero.
Here is a <characteristic function> proof using only the finite second moment. Put $U=(X_1-\mu)/\sigma$, so $\mathbb EU=0$ and $\mathbb EU^2=1$. The elementary integral remainder identity gives $|e^{iz}-1-iz|\leq z^2/2$ for real $z$. Consequently
$$
\frac{e^{itU}-1-itU}{t^2}\longrightarrow-\frac{U^2}{2},\qquad
\left|\frac{e^{itU}-1-itU}{t^2}\right|\leq\frac{U^2}{2}.
$$
By the <dominated convergence theorem>, the <characteristic function> satisfies $\varphi_U(t)=1-t^2/2+o(t^2)$ as $t\to0$. For a fixed real $t$, <independence> therefore gives
$$
\varphi_{(U_1+\cdots+U_n)/\sqrt n}(t)
=\varphi_U(t/\sqrt n)^n
=\left(1+\frac{c_n}{n}\right)^n,\qquad c_n\longrightarrow-t^2/2.
$$
The allowed complex-number limit makes this tend to $e^{-t^2/2}$, the <characteristic function> of the standard <normal distribution>. It is continuous at zero, so the <Lévy continuity theorem> proves the asserted <convergence in distribution>. No third moment or boundedness of the summands was used.
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