If $|q|<1$ and we set
$$
(a;q)_{\infty}=\prod^{\infty}_{n=0}\left(1-aq^n\right),
$$
then (see [1]):
$$
P(a,q):=\left(\frac{(-a;q)_{\infty}}{(a;q)_{\infty}}\right)^2=
$$
$$
=-1+\frac{2}{1-}\frac{2a}{1-q+}\frac{a^2(1+q)^2}{1-q^3+}\frac{a^2q(1+q^2)^2}{1-q^5+}\frac{a^2q^2(1+q^3)^2}{1-q^7+}\ldots,\tag 1
$$
where $a$ is a complex number.

Take the logarithms in (1) and expand the two products in Taylor series (of $\log(1-x)$). Then the double sums are easily rearranged and we get
$$
4\sum^{\infty}_{n=0}\frac{a^{2n+1}}{(2n+1)(1-q^{2n+1})}=\log P(a,q). \tag 2
$$
Set also
$$
u_0(a,q):=\frac{P(a,q)-1}{P(a,q)+1}. \tag 3
$$
Then we have
$$
u_0(a,q):=\frac{2a}{1-q+}\frac{a^2(1+q)^2}{1-q^3+}\frac{a^2q(1+q^2)^2}{1-q^5+}\frac{a^2q^2(1+q^3)^2}{1-q^7+}\ldots,\tag 4
$$
Set now the value $a=q^{\nu+1/2}$, $\nu\in\{0,1,2,\ldots\}$ in (2) to get
$$
P\left(q^{\nu+1/2},q\right)=\exp\left(-4\sum^{\infty}_{n=0}\frac{q^{(2n+1)(\nu+1/2)}}{(2n+1)(1-q^{2n+1})}\right)=
$$
$$
=\exp\left(-4\sum^{\nu-1}_{j=0}\textrm{arctanh}(q^{j+1/2})+\textrm{arctanh}(k_r)\right) \tag 5
$$

( For many applications and very interested notes see:

http://www-elsa.physik.uni-bonn.de/~dieckman/InfProd/InfProd.html )

Here the function $k_r$ is the elliptic singular modulus and $k'_r=\sqrt{1-k_r^2}$.

For $\nu=0$ in (5) we get
$$
P\left(q^{1/2},q\right)=\frac{k'_r}{1-k_r}.
$$
Hence your continued fraction $u_0(q^{1/2},q)$ is
$$
u_0(q^{1/2},q)=\frac{k'_r+k_r-1}{k'_r-(k_r-1)}=\frac{2q^{1/2}}{1-q+}\frac{q(1+q)^2}{1-q^3+}\frac{q^2(1+q^2)^2}{1-q^5+}\frac{q^3(1+q^3)^2}{1-q^7+}\ldots,\tag 6
$$
where $q=e^{-\pi\sqrt{r}}$, $r>0$.

Relation (6) with $r=1$, $k_1=k'_1=\frac{1}{\sqrt{2}}$ gives:
$$
u_0\left(e^{-\pi/2},e^{-\pi}\right)=\sqrt{2}-1.
$$

References

[1]: N.D. Bagis and M.L. Glasser. "Evaluations of a Continued Fraction of Ramanujan". Rend. Sem. Mat. Univ. Padova, Vol. 133 (2015). (submited 2013)