Let $x_1,\dots,x_n$ be i.i.d. drawn from $N(0,I_{p\times p})$. Let $\hat S = \frac 1n \sum_{i=1}^n x_ix_i^T$ be the sample covariance. Let $\lambda_1 \ge ... \ge \lambda_p$ be the eigenvalues of $\hat S$.

It is known that $$\mathbb E\frac1p\operatorname{tr}\left(\hat S^{-1}\right) = \mathbb E\frac1p\sum_{i=1}^p \frac{1}{\lambda_i} = \frac{n}{n-p-1},$$ Therefore, if $p=n$ this quantity tends to $\infty$ as $n,p\to \infty$.

Suppose that $\beta>0$ and consider the sum $$\frac1p\sum_{i=1}^p \frac{1}{\lambda_i+\beta}.$$

Given any number $a>0$, is there a $\beta$ such that as $n\to\infty $ and $p/n\to 1$ $$\mathbb E\frac1p\sum_{i=1}^p \frac{1}{\lambda_i+\beta}\to a?$$ This looks like it should be straightforward, but the analysis appears subtle because of the behavior of the least eigenvalues of $\hat S$. It doesn't seem quite enough to show that for any $\beta>0$ this sum is positive and monotone increasing, because of the question of how the eigenvalues behave in the limit.

Any help would be greatly appreciated.