Thank you for the enlightening answer. I am trying to see what best can be said about $\tilde{E}$. Neverthless $\tilde{E}$ is not well defined, irrespective of how we choose the eigenvector matrix $\tilde{E}$, can we always say the following : Let $\Sigma$ is any constant diagonal matrix, as $\epsilon\to 0$, $\tilde{E}\Sigma\tilde{E}^T$ always converges to some matrix.

The crucial for the answer is the value $s$ and it is not the order of singularity of $A$. It depends on first $p$ matrices in the series and there is a procedure to determine $s$. pdfs.semanticscholar.org/106b/…

Also clearly the statement in that paper that $H_{-s} \ne 0$ is wrong as per your example $A+I/{\lambda}$ when $n = 3$.

In some of my computations too I observed the same. I get $\lambda$ increase and not any higher powers.

Oh the paper clearly says $H_{-s}$ is not zero. So thats about it?

Yes, that's something need to be resolved. To see if it can make the leading $n-2$ matrices zero.

@losifPinelis : Thanks for the example. Looks like the eigenvectors should converge elementwise.

cross post : math.stackexchange.com/q/3775350/2987

@vidyarthi : I feel its better to write all steps or leave it like that.