Consider a series of random matrices $X_n\in\mathbb{R}^{n\times m}$ consisting of i.i.d. entries, each with zero mean and variance $1/m$, and let $a_n,b_n\in\mathbb{R}^{n\times1}$ be two deterministic (or random and independent on $X$) vectors, say with bounded norm.

I want to find the structure of some "nice"/"simple" "limit" function, $f_n$, of the following term $$ a_n^T\left(X_nX_n^T+I_n\right)^{-1}b_n-f_n\to0 $$ almost surely, as $n,m\to\infty$ with fixed ratio.

**EDIT**: Due to Ofer's answer and comments I will consider some specific choice of $a_n$:
$$
\frac{1}{n}w_n^TX_n^T\left(X_nX_n^T+I_n\right)^{-1}b_n-f_n\to0
$$
Since $w_n^TX_n^T = \sum_{i=1}^mw_{i}x_i^T$ where $x_i$ is $i$th raw of $X_n$, we can write that
\begin{align}
\frac{1}{n}w_n^TX_n^T\left(X_nX_n^T+I_n\right)^{-1}b_n = \frac{1}{n}\sum_{i=1}^mw_{i}x_i^T\left(X_nX_n^T+I_n\right)^{-1}b_n
\end{align}
We know that $X_nX_n^T = \sum_{i=1}^mx_ix_i^T$. Let $\left[X_nX_n^T\right]_i = X_nX_n^T-x_ix_i^T$. Thus,
\begin{align}
\frac{1}{n}w_n^TX_n^T\left(X_nX_n^T+I_n\right)^{-1}b_n &= \frac{1}{n}\sum_{i=1}^mw_{i}x_i^T\left(X_nX_n^T+I_n\right)^{-1}b_n\\
&=\frac{1}{n}\sum_{i=1}^mw_{i}\frac{x_i^T\left(\left[X_nX_n^T\right]_i+I_n\right)^{-1}b_n}{1+x_i^T\left(\left[X_nX_n^T\right]_i+I_n\right)^{-1}x_i}
\end{align}
Now, since $x_i$ is independent on $\left(\left[X_nX_n^T\right]_i+I_n\right)^{-1}$, we know that (a.s.)
$$
x_i^T\left(\left[X_nX_n^T\right]_i+I_n\right)^{-1}x_i-\int (1+x)^{-1} \rho(dx)\to0
$$
where $\rho$ is the limit density of eigenvalues of $XX^T$. The same is true for the numerator. So, the speculation is that $f_n$ behaves like
\begin{align}
f_n &= \frac{1}{n}\sum_{i=1}^mw_{i}\frac{x_i^Tb_n\int (1+x)^{-1} \rho(dx)}{1+\int (1+x)^{-1} \rho(dx)}\\
&=\frac{\int (1+x)^{-1} \rho(dx)}{1+\int (1+x)^{-1} \rho(dx)}\frac{1}{n}w_n^TX_n^Tb_n
\end{align}

**Update**: Numerical calculations suggests that the above "limit" is not true, although, I can't really say that I completely understand where the rub is.