Central limit theorem

Let $T$ be a zero-mean unit-variance random variable with probability density function $f(t)$. If its variance is not $1$ then divide $T$ by its standard deviation. We show that $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(T_1+\dots+T_n\right)$$ approaches standard normal random variable, where $T_1,\dots,T_n$ are independent samples from $T$.

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Proof:

Denote by $F(\mu)$ the Fourier transform of $f(t)$, written as $$f(t)\leftrightarrow F(\mu).$$ $f(t)$ is a probability density function implies that $F(0)=1$. Given $$tf(t)\leftrightarrow \frac{jF'(\mu)}{2\pi}$$ and $$\int_{-\infty}^\infty tf(t)dt=0$$ we have $F'(0)=0$. Since $$\int_{-\infty}^\infty t^2f(t)dt=1$$ and $$t^2f(t)\leftrightarrow-\frac{1}{4\pi^2}F''(\mu),$$ we get $F''(0)=-4\pi^2$. Hence we write $$F(\mu)=F(0)+\mu F'(0)+\frac{\mu^2}{2}F''(0)+\dots=1-2\pi^2\mu^2+\frac{F'''(0)}{6}\mu^3+\cdots.$$Adding $n$ independent samples of $T$ translates to convolution of $n$ copies of $f(t)$, which translates to $F(\mu)^n$. Division by $\sqrt{n}$ translates to replacing $\mu$ with $\mu/\sqrt{n}$. Thus the Fourier transform of the probability density function of interest is $$F\left(\frac{\mu}{\sqrt{n}}\right)^n=\left(1-\frac{2\pi^2\mu^2}{n}+\frac{F'''(0)}{6n^{1.5}}\mu^3+\cdots\right)^n\approx e^{-2\pi^2\mu^2},$$which is the Fourier transform of $$\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}.$$