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.
 A: OP's underlying question is essentially the same as this Phys.SE post, although the detailed calculation is slightly different and interesting to compare.
I) The action for a free non-relativistic point particle with mass $m=1$ reads:
$$\tag{1} S ~=~\frac{1}{2}\int_0^T\! dt~ \dot{x}(t)^2~=~ \frac{1}{2}\langle x,Ax \rangle~=~\frac{1}{2}\sum_{n\in \mathbb{N}} \lambda_n c_n^2 . $$
Here we have assumed Dirichlet boundary conditions (DBC)
$$\tag{2} x(0)~=~0~=~x(T).$$
Moreover, here
$$\tag{3}  \langle f,g  \rangle ~:=~ \int_0^T\! dt ~f(t)g(t) $$
is an inner product over $\mathbb{R}$. 
II) In eq. (1) we have also introduced a positive operator
$$\tag{4} A~:=~-\partial_t^2 $$
with positive eigenvalues 
$$\tag{5} \lambda_n~=~\left(\frac{\pi n }{T}\right)^2~>~0,  \qquad n\in \mathbb{N}.$$
The determinant becomes via zeta-function regularization
$$\tag{6}\det(A)~=~\prod_{n\in\mathbb{N}} \lambda_n~=~\left(\prod_{n\in\mathbb{N}} \frac{\pi n }{T}\right)^2
~=~2T ,$$
using e.g. eq. (7) in my Phys.SE answer here.
III) The normalized eigenfunctions are  
$$\tag{7} x_n(t) ~=~ \sqrt{\frac{2}{T}} \sin \frac{\pi n }{T}t , \qquad n\in \mathbb{N}. $$   
An arbitrary virtual path $t\mapsto x(t)$ that satisfies the DBC (2) is a linear combination
$$\tag{8}  x ~=~ \sum_{n\in \mathbb{N}} c_n x_n, $$
where $c_n\in\mathbb{R}$ are arbitrary coefficients, which we should integrate over in the path integral.
IV) Now let us consider quantum mechanics. Let us assume $\hbar=1$ for simplicity. The path integral measure is
$$\tag{9} {\cal D}x~:=~N \prod_{n\in\mathbb{N}}   \frac{\mathrm{d}c_n}{\sqrt{2\pi}} ,  $$
where $N$ is a normalization factor. So the Euclidean path integral is an infinite-dimensional Gaussian integral
$$\tag{10} Z~=~\int_{DBC} \!{\cal D}x ~e^{-S}~=~\frac{N}{\sqrt{\det (A)}} 
~=~ \frac{N}{\sqrt{2T}}. $$
Apparently we should chose the normalization factor $N=\frac{1}{\sqrt{\pi}}$ in order to achieve the Euclidean version of OP's first formula
$$ \tag{11} Z~=~\frac{1}{\sqrt{2 \pi T}}.$$
