The thermodynamic expectation value for an observable $A$ is defined as $$\langle A \rangle = \frac{1}{Z} \sum_n \langle\psi_n| e^{-\beta H} A|\psi_n \rangle, \qquad (1)$$ where $\beta=1/k_bT$, the $\psi_n$ are a basis for the Hilbert space and $$Z= \sum_n \langle\psi_n| e^{-\beta H} |\psi_n\rangle. $$ Now, in the limit $T\rightarrow 0$ (or $\beta \rightarrow \infty$), only the ground state should contribute, hence I would expect that $$\langle A \rangle = \langle A \rangle_0 = \langle \psi_0 | A |\psi_0 \rangle,$$ where $\psi_0$ is the ground state.
What I want to know is, whether this is correct and if it is correct, how can I proof this, starting from eq. (1). I started with assuming that the $\psi_n$ are the energy eigenstates of the Hamiltonian $H$ (with $\psi_0$ being the eigenstate corresponding to the lowest energy $E_0$ [with $E_0<E_1<E_2\ldots$ ]) and rewrote the expectation value as $$\langle A \rangle = \frac{ \langle A \rangle_0 + \sum_{n=1}^N e^{-\beta (E_n-E_0)} \cdot \langle \psi_n| A|\psi_n \rangle}{\sum_{n=0}^N e^{-\beta (E_n-E_0)}}.$$
But from here I cannot see what happens, if I take $\beta\rightarrow \infty$.