Let $X$ be a separable Hilbert space, and let $(e_i)_{i=1}^\infty$ be an orthonormal basis of $X$. For each $n\in \mathbb{N}$, let $X_n$ be  the subspace  spanned by $(e_i)_{i=1}^n$, and consider the operator $I_n:\mathbb{R}^n\to X$ be such that 
$I_n u=\sum_{i=1}^n u_i e_i$ for all $u=(u_i)_{i=1}^n$.  The adjoint $I_n^*:X\to \mathbb{R}^n$ is then given by 
$I^*_n x= (\langle x, e_i\rangle_X)_{i=1}^n$. 

Let $M$ be a symmetric positive semidefinite operator on $X$ (we can assume $M$ is of trace class if necessary). 
Given $\lambda>0$ and $U\in X$. Is it true that 
$$
\lim_{n\to \infty} \langle I_n(I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n})^{-1} I_n^* U, U\rangle_X
 = \langle  (  M +\lambda \operatorname{id}_{X})^{-1}   U,U\rangle_X.
$$


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Thanks to Prof. Ozawa's comment, here is what I have got so far: 

As $B= I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n}$ is a symmetric positive definite matrix on $\mathbb{R}^n$, 
$I_nBI_n^*$ preserves the space $X_n$, and hence
one can directly verify that 
$$
 I_nB^{-1} I_n^*=P_n (I_nBI_n^*)^{-1} P_n,
$$
where $P_n$ is the orthogonal projection onto $X_n$.
As $I_nI_n^*=P_n =P_n  \operatorname{id}_{X} P_n$,
\begin{align*}
I_n(I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n})^{-1} I_n^* 
&=
P_n (I_n(I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n}) I_n^*)^{-1} P_n  
\\
&=P_n (P_n(  M +\lambda \operatorname{id}_{X}) P_n)^{-1} P_n.
\end{align*}
Then the desired claim reduces to   the  convergence of 
$$
P_n (P_n(  M +\lambda \operatorname{id}_{X}) P_n)^{-1} P_n
\to (  M +\lambda \operatorname{id}_{X})^{-1}
$$
 in the weak operator topology. 
However,  I don't know how to handle the inverse operator.  In particular,   $P_n(  M +\lambda \operatorname{id}_{X}) P_n$ is only invertible on $X_n$, $(P_n(  M +\lambda \operatorname{id}_{X}) P_n)^{-1}$ is only defined on $X_n$.