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Suppose $X_n\to X$ with convergence in probability. For real $t$ and $\delta >0$, the elementary inequality $|e^{iu}-e^{iv}|\le \min (2,|u-v|)$ givesFirst let $n\to \infty$, then let $\delta \downarrow 0$. The characteristic functions converge pointwise to that of $X$, which is continuous at zero. The Lévy continuity theorem gives convergence in probability implies convergence in distribution.

For the converse, let $Z$ have the fair Bernoulli distribution, put $X=Z$, and set $X_n=1-Z$ for every $n$. All $X_n$ have the same probability distribution as $X$, so $X_n\stackrel{d}{\to }X$. But $|X_n-X|=1$ always, henceThere is no convergence in probability to $X$. In this example $X_n$ does have convergence in probability to $1-Z$: the counterexample concerns the converse with the specified limit $X$, as usual. If one interprets the question as denying convergence in probability to any limit, take $X_n=Z$ for even $n$ and $X_n=1-Z$ for odd $n$. Its probability distribution is still constant, while the two subsequences have distinct limits from almost sure convergence, ruling out any common limit in probability by uniqueness of a limit in probability.

Since $Y_1=X_1^2$ and a standard normal variable has second and fourth moments $1$ and $3$,Thus $T_n$ has the chi-squared distribution with $n$ degrees of freedom and

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 giveUsing $\log (1-z)=-z-z^2/2+O(z^3)$,Therefore ${\phi }_{V_n}(t)\to e^{-t^2/2}$, the characteristic function of the standard normal distribution. Lévy's theorem now givesThis derives the limit directly, without assuming the central limit theorem.

A finite linear combination of independent Gaussian random variables is Gaussian, soSince $v_n\to v:={\sum }_{k\ge 1}{a}_k^2<\infty$, the characteristic functions converge pointwise:The Lévy continuity theorem yields convergence in distribution toOn the given common probability space, the same variance calculation actually proves convergence in $L^2$ as well.

Differentiability gives $\phi (u)=1+iau+o(u)$ at zero. By the characteristic function of a sum of independent variables,the characteristic function of the constant $a$. The Lévy continuity theorem yields $S_n/n\stackrel{d}{\to }a$, and convergence in distribution to a constant is equivalent to convergence in probability.