# Inverse of a matrix with main diagonal elements approaching infinity

Let $A$ be a invertible, symmetric, positive definite $p \times p$ covariance matrix with main diagonal elements $a_{ii},~i = 1,~\ldots,~p$. If all main diagonal elements would approach $\infty$, would the main diagonal elements of $A^{-1}$ approach $0$?

Two month ago on Crossvalidated I asked whether $(X'X)^{-1} \to 0$ as $n \to \infty$ (https://stats.stackexchange.com/questions/208416/decrease-of-xx-1-as-n-increases). There I hoped for some statistical hints, as a consistent estimator can have the variance $(X'X)^{-1}$. While trying to find out whether the variance of such an estimator always become $0$ I found https://stats.stackexchange.com/questions/74047/why-dont-asymptotically-consistent-estimators-have-zero-variance-at-infinity, pointing in the direction that this doesn't always have to be the case, but the cited exam in Consistent Estimator and Convergence Variance states that it is true.

Now I want to try a more mathematical approach. A question slightly related is https://mathoverflow.net/questions/224061/inverse-of-matrix-with-one-element-approches-infinity, but I want to have all main diagonal elements to approach infinity (and also I think the Sherman–Morrison formula doesn't help as it would need the inverse of a matrix with infinite large elements). Literature that might be helpful (but I didn't see something answering my question) is http://www.sciencedirect.com/science/article/pii/S0024379508004618 and http://www.sciencedirect.com/science/article/pii/0024379587903417.

Edit: I changed the question to $A$ being a symmetric, positive definite covariance matrix.

No, not necessarily. Consider $p=2$ and $$A=\begin{pmatrix} n & n+1 \cr n-1 & n \end{pmatrix},$$ for $n\to \infty$. Then we have $$A^{-1}=\begin{pmatrix} n & -n-1 \cr -n+1 & n \end{pmatrix},$$ but still the main diagonals go to $\infty$.
Edit: OK, the question was changed. $A$ now has to be a symmetric, positive definite $p \times p$ covariance matrix. Then the example has to be modified: $$A=\begin{pmatrix} n & n \cr n & n+\frac{1}{n} \end{pmatrix}$$ It has determinant $1$, is symmetric and positive definite.