= Solution
Since $Y_1=X_1^2$ and a standard normal variable has second and fourth moments $1$ and $3$,
$$
\boxed{\mu=\mathbb EY_1=1,
\qquad \sigma^2=\operatorname{var}(Y_1)=3-1=2}.
$$
Thus $T_n$ has the <chi-squared distribution> with $n$ degrees of freedom and
$$
V_n=\frac{T_n-n}{\sqrt{2n}}.
$$
<Convergence in distribution> $W_n\Rightarrow W$ means that the <cumulative distribution functions> satisfy $F_{W_n}(x)\to F_W(x)$ at every continuity point of $F_W$. The <Lévy continuity theorem> states that this is equivalent to pointwise convergence of <characteristic functions> to the characteristic function of $W$, provided the limiting function is continuous at zero.
For fixed $t$, independence and a Gaussian integral give
$$
\varphi_{V_n}(t)
=\left[e^{-it/\sqrt{2n}}
\left(1-i t\sqrt{\frac2n}\right)^{-1/2}\right]^n.
$$
Using $\log(1-z)=-z-z^2/2+O(z^3)$,
$$
\log\varphi_{V_n}(t)
=n\left[-\frac{it}{\sqrt{2n}}
-\frac12\log\left(1-it\sqrt{\frac2n}\right)\right]
=-\frac{t^2}{2}+O(n^{-1/2}).
$$
Therefore $\varphi_{V_n}(t)\to e^{-t^2/2}$, the characteristic function of the <standard normal distribution>. Lévy's theorem now gives
$$
\boxed{V_n\Rightarrow Z}.
$$
This derives the limit directly, without assuming the central limit theorem.
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