quantum mechanics - Free space propagator: reconciling two results

In quantum mechanics, the free space propagator $G(q_f=0,q_i=0;\tau)$ can be easily calculated to be $$\sqrt{\frac{m}{2\pi i \hbar \tau}}$$ by inserting an identity operator.

However if we use functional integral, we get $$\begin{equation} \begin{split} G(q_f=0,q_i=0;\tau)&=\int Dq e^{-\frac{i}{\hbar}\int_0^{\tau} dt \frac{m}{2}\dot{q}^2}\\ &=\int Dr e^{-\frac{i}{\hbar}(S[q_{cl}]+S[r(t)])}\\ &=\int Dr e^{-\frac{i}{\hbar}S[r(t)]}\\ &=\int Dr e^{\frac{i}{\hbar}\int dt r(t)\partial_t^2 r(t)}\\ &=(\det[\frac{i}{\pi\hbar}\partial_t^2])^{-1/2} \end{split} \end{equation}$$ where the classical trajectory $q_{cl}(t)=0$ due to the boundary conditions and $r(t)$ is the fluctuation. If we solve for the eigenstates and eigenvalues of $\partial_t^2$: $$\partial_t^2 r_n(t)=\lambda_nr_n(t)$$ with $r_n(0)=r_0(\tau)=0$, we get $r_n(t)=\sin(n\pi t/\tau)$ and $\lambda_n=(n\pi/\tau)^2$. Therefore, we have $$\det(\partial_t^2)=\prod_{j=1}^{\infty}(n\pi/\tau)^2$$ which goes to infinity and as a result the propagator seems to go to 0.

I'm not sure where went wrong for this calculation. Any help is appreciated.

Edit: suggested by @AccidentalFourierTransform, below is the zeta function approach, which still doesn't seem to work.

for simplicity we set all the irrelevant constants to 1, and thus $\lambda_n=n^2$, then we have $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{\lambda_n^s}=\sum_{n=1}^{\infty}\frac{1}{n^{2s}}$$ and then we need to calculate the derivative of the zeta function and then taking the limit of $s$ goes to 0 followed by exponentiation in order to obtain the determinant.

$$\zeta'(s)=\sum_{n=1}^{\infty}-\frac{ln\lambda_n}{n^{2s}}$$ I tried numerically by taking $s$ to 0 both from the real axis and the imaginary axis, but both seems to diverge, i.e the same problem remains.