I am trying to evaluate this limit:
$$\lim_{n \to \infty} \left[(\text{det}\ \Gamma_n)^{-\frac{1}{2}}\exp \left\{ {-\frac{1}{2}(x-m_n)\cdot (\Gamma_n)^{-1}(x-m_n)} \right\} \right]$$
where $\Gamma_n$ is a singular, $k \times k$ matrix, and $x$ and $m_n$ are $k$ dimensional vectors, such that $\lim\limits_{n \to \infty} \Gamma_n = \Gamma$ and $\lim\limits_{n \to \infty} m_n = m$. I know that the limit is supposed to be equal to
$$\left[(\text{det}\ \Gamma)^{-\frac{1}{2}}\exp \left\{ {-\frac{1}{2}(x-m)\cdot (\Gamma)^{-1}(x-m)} \right\} \right]$$
but I am having a really hard time showing this, I'm not sure how to work with limits in $k$ dimensions, involving vectors and matrices, like this.
Note: this problem actually asks to show that if $m_n \to m$ and $\Gamma_n \to \Gamma$, where $m_i$ is the mean of a Gaussian probability distribution $P_i$ on $R^k$ and $\Gamma_i$ is the covariance matrix of that distribution, that $P_n$ converges in distribution to $P$, where $P$ is a Gaussian with mean vector $m$ and covariance matrix $\Gamma$. I am trying to do this problem by using the Cramer-Levy continuity theorem, and thus by showing that the characteristic functions of the $P_i$ converge... if there is a better way to do this, I would appreciate seeing it!
The assertion $\Gamma_n\to\Gamma$ means that each entry of $\Gamma_n$ converges to the corresponding one of $\Gamma$. Since the determinant of a matrix is a continuous functions of its entries, we have $\det\Gamma_n\to\det\Gamma$, hence, by continuity of the exponential function, it remains to prove that for each $x\in\mathbf R^k$, $$\lim_{n\to \infty} (x-m_n)^T\cdot (\Gamma_n)^{-1}(x-m_n) =(x-m)^T\cdot \Gamma^{-1}(x-m).$$ This can be done by writing for a symmetric matrix $A$ the quantity $y^tAy$ in terms of the entries of $A$ and that of $y$. Note that using the formula of the inverse of matrix with the comatrix, we have $\Gamma_n^{-1}\to \Gamma^{-1}$.