Calculating the partition function for a single mode free boson gas using path integrals Considering a single free bosonic mode with Hamiltonian $H=\epsilon a^{\dagger}a$, we can use (bosonic) coherent states to write the corresponding partition function $\mathcal{Z}=\mathrm{tr} (e^{-\beta H})$ as:
\begin{equation}
\mathcal{Z}=\int d(\bar{\psi},\psi)e^{-\bar{\psi}\psi}\langle\psi|e^{-\beta H}|\psi\rangle,
\end{equation}
where we integrate over the complex variables $\bar{\psi}$ and $\psi$. To solve this using a path integral we apply the usual procedure, partitioning the imaginary time in $N$ intervals and subsequently inserting resolutions of identity. This simply yields the usual path integral expression (cf. eq. (4.27) in A. Altland's "Condensed matter field theory"):
\begin{equation}
\mathcal{Z}=\int \prod_i d(\bar{\psi}_i,\psi_i)e^{\sum_i-\bar{\psi}_i\psi_i + \bar{\psi}_{i+1}\psi_i-\delta\epsilon\bar{\psi}_{i+1}\psi_i},
\end{equation}
where $\delta\equiv \beta/N$. Usually, one would take the continuum limit at this point, but instead I'd like to first evaluate this integral for large $N$, and afterwards take $N\rightarrow \infty$ (since the integral is quadratic, this should be fine). This being a homework problem, there is a hint that says:
\begin{equation}
 1+a+a^2+a^3+...=\frac{1}{1-a},
\end{equation}
which seems to indicate that the $\mathcal{Z}$ above should be evaluated at each seperate timeslice, resulting in 
\begin{equation}
\mathcal{Z}=\sum_n e^{-\beta n\epsilon},
\end{equation}
since we know from quantum statistical mechanics that a single bosonic mode has partition function
\begin{equation}
\frac{1}{1-e^{-\beta \epsilon}}.
\end{equation}
In trying to carry out this program, I have tried to use the standard complex Gaussian, but in no way do I seem to get the series mentioned in the hint. Therefore, my question is: how do I evaluate $\mathcal Z$ above for large $N$? 
 A: To begin with, the result of the multi-dimensional complex Gaussian integral is
\begin{equation}
I(A) := \int d(z_1, z_1^\ast) \cdots d(z_n, z_n^\ast)\, \exp\big(-\vec{z}^{\dagger} A \vec{z}\big) = \det(A)^{-1}.
\end{equation}
Here, $A$ is an $n\times n$ matrix, and $\vec{z} = (z_1, \ldots, z_n)^T$. The integration measure $d(z, z^\ast)$ is defined by
\begin{equation}
d(z, z^\ast) := \frac{d\mathrm{Re}(z)\, d\mathrm{Im}(z)}{\pi}.
\end{equation}
Then, the partition function of a single free bosonic mode with $H = \epsilon \,a^\dagger a$ is simply
\begin{equation}
\mathcal{Z} = \lim_{n \to \infty} I(M_n) = \lim_{n \to \infty}\det(M_n)^{-1}
\end{equation}
with
\begin{equation}
M_n := \begin{pmatrix}1&&&&-1+x_n\\ -1+x_n&1&&&\\& -1+x_n&1&&\\ &&\ddots&\ddots&\\ &&&-1+x_n&1\end{pmatrix},
\end{equation}
where $x_n := \beta\epsilon/n$. From the determinant formula
\begin{equation}
\det(A) = \varepsilon_{j_1 \ldots j_n} A_{1j_1}A_{2j_2} \ldots A_{nj_n},
\end{equation}
it is easy to see that $\det(M_n) = 1 - (1-x_n)^n$. Then, we have
\begin{equation}
\lim_{n\to \infty} (1 - x_n)^n = \left(1-\frac{\beta\epsilon}{n}\right)^n = e^{-\beta\epsilon},
\end{equation}
from which it follows that
\begin{equation}
\mathcal{Z} = \lim_{n \to \infty}\det(M_n)^{-1} = (1 - e^{-\beta\epsilon})^{-1}.
\end{equation}
A: Do you need to use the coherent states to calculate the partition function? You might be overthinking it a bit.
A partition function is the sum of Boltzmann factors ($e^{-\beta E}$) over the Fock space. Your Fock space is defined by the occupation numbers: $|0\rangle$, $|1\rangle$, etc. For you, $\hat H|n\rangle = n\epsilon|n\rangle$. So you need to perform the sum as follows:
$\mathcal{Z} = \sum_n \langle n|e^{-\beta \hat H}|n\rangle = \sum_n e^{-\beta n\epsilon}$,
which is exactly the formula from the hint.
If you insist on using QFT, start by writing down the action for your system using Matsubara frequencies:
$S[\bar\psi,\psi] = \sum_n \bar\psi_n\left[-i\omega_n + \epsilon\right]\psi_n$
Your partition function is
$\mathcal{Z} = \int D(\bar\psi,\psi) e^{-S[\bar\psi,\psi]}$. 
From your question, I assume you have Altland's book, so to proceed see Eq.(4.33)
