Consider a sequence of matrices with the following properties:
The diagonal elements approach positive infinity as n increases.
The other elements are constants.
In this case, will the inverse of these matrices approach the 0 matrix as n increases? If not, will the diagonal elements in the inverse approach 0?
According to the OP description, the sequence is $D_n+A$, where $D_n$ is a diagonal matrix with elements on the diagonal going to $\infty$, and $A$ is constant. Factoring out, $(D_n+A)^{-1}=D_n^{-1}(I+D_n^{-1}A)^{-1}$.
Suppose $d_n$ is the smallest diagonal element of $D_n$, then $||D_n^{-1}A||\leq d_n^{-1}||A||\to0$. So for large $n$ we have $||D_n^{-1}A||<1$, and according to the Neumann series for the inverse: $$(I+D_n^{-1}A)^{-1}=I-D_n^{-1}A+(D_n^{-1}A)^2-(D_n^{-1}A)^3+\dots$$
Since, by summing the convergent geometric series, $$||I-D_n^{-1}A+(D_n^{-1}A)^2-(D_n^{-1}A)^3+\dots||\leq 1+||D_n^{-1}A||+||(D_n^{-1}A)||^2+||(D_n^{-1}A)||^3+\dots=\frac1{1-||D_n^{-1}A||},$$
we have that $$||(D_n+A)^{-1}||\leq \frac{||D_n^{-1}||}{1-||D_n^{-1}A||}\to0.$$ So yes, the inverses do converge to $0$.
