Proof that the inverse of a growing matrix whose elements converge to the identity matrix converges to the identity matrix

Question

Consider a matrix of the form $I+A$, where $I$ is an identity matrix of size $n$ and $A$ is a matrix of size $n$ where each element is less that $\tfrac{1}{n} B $ in absolute value ($B>0$). I want to prove that the elements of the inverse $(I+A)^{-1}$ converge to the elements of the identity matrix of size $n$ as $n \to \infty$. The bound $B$ does not change as the matrix size increases. I can assume that $A$ is symmetric positive definite if that helps. I can't assume that the size of the elements converge to zero at a rate faster than $\tfrac{1}{n}$. This result seemed to be true in simulations I ran, but I have been stuck on coming up with a proof.

Answer 1

This is false. If $A$ is $n\times n$ matrix with each element equal to $-1/n$, then $I+A$ is not invertible.

Answer 2

I believe this is a true assertion when $B<1$, and here is a possible proof: each individual matrix $A^k$ tends to the zero matrix as $n\to\infty$, and we have the series $$ (I+A)^{-1} = I - A + A^2 - A^3 + A^4 - \cdots. $$ One would have to justify taking the limit term by term as $n\to\infty$.