Start with the circle described by
$$\left(x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}\right)$$
to get
$$\frac{\pi}{4} = \int_0^1\frac{dt}{1+t^2} = \frac12\int_0^1\frac{s^{-1/2}}{1+s}\,ds$$
using elementary calculus.
Now define
$$G(z)\;:=\;\frac12\int_0^1\frac{s^{-1/2}}{1+z s}\,ds =\frac12\sum_{n\ge0}(-z)^n\int_0^1 s^{n-1/2}\,ds =\sum_{n\ge0}\frac{(-z)^n}{2n+1}$$
and note that using $z=1$ gives
$$\boxed{\sum_{n\ge0}\frac{(-1)^n}{2n+1}= \pi/4}$$
which is the famous Madhava-Gregory-Leibnitz-(enter your favourite mathematician) series for $\pi$. Squaring both sides of this "carefully", one gets
$$\boxed{\sum_{n\ge 1} \frac 1{n^2} = \frac{\pi^2}{6}}$$
which is a famous result of Euler. This "careful squaring" is an exercise in Borwein & Borwein’s Pi and the AGM (Wiley, 1987).
Now, consider the character $\chi$ mod $4$ and define
$$F(s) := \frac{1}{1 - 2^{-s}} \cdot \prod_{p\in \mathcal P} \frac{1}{1 + \frac{\chi(p)}{p^s}}$$
where $\mathcal P$ is the set of odd primes.
A little algebra shows
$$F(s) = \frac{\zeta(2s)(1 + 2^{-s})}{L(s, \chi)}$$
where
$$\zeta(s) := \sum_{n\ge 1} \frac{1}{n^s}, \qquad L(s, \chi) := \sum_{n\ge 1} \frac{\chi(n)}{n^s} = \prod_{p} \frac{1}{1 - \frac{\chi(p)}{p^s}}$$
for our chosen character.
Taking $s\to 1^+$ proves
$$\boxed{\pi = \sum_{m=1}^{\infty}\frac{(-1)^{s(m)}}{m}}$$
where $s(m)$ counts the number of appearances of primes of the form $4k+1$ in the prime decomposition of $m$.
We return back to $G$ and note that $G$ can also be expressed as the Stieltjes integral
$$G(z)=\int_0^1\frac{d\mu(s)}{1+zs}$$
for the positive measure $\mu$ on $[0,1]$ with density $d\mu(s)=\tfrac12 s^{-1/2}ds$. Now, show that any function of this type admits an expression of the form
$$G(z)=\cfrac{1}{\,1+\cfrac{\beta_1 z}{3+\cfrac{\beta_2 z}{5+\cfrac{\beta_3 z}{7+\ddots}}}\,}$$
by looking at the Taylor series and uniquely determining the finite continued fraction convergents successively so that the $k$-th convergent matches the Taylor series up to $z^k$.
The crucial part is to show $\beta_n = n^2$. If we can prove this, then we get
$$\boxed{\frac{\pi}{4} = \cfrac{1}{\,1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\ddots}}}\,}}$$
from which one also gets
$$\boxed{\frac{\pi}{4}=1+ \cfrac{1}{3- \cfrac{3\cdot 4}{1- \cfrac{2 \cdot 3}{3- \cfrac{5 \cdot 6}{1-\cfrac{4 \cdot 5}{\ddots}}}}}}$$
after some simple algebra.
Indeed, start with the first form and write
$$\begin{align*}\frac 4{\pi} -1 &= \frac{1}{3+\dfrac{4}{5+\dfrac{9}{7+\cdots}}} = \frac{1}{3+\dfrac{4}{\,\dfrac{\left(5+\dfrac{9}{7+\cdots}\right)}{1}\,}} = \frac{1}{\,3+\dfrac{4}{\dfrac{5+\dfrac{9}{7+\cdots}}{1}\,}}\\&= \frac{1}{\,3+\dfrac{4}{5+\dfrac{9}{7+\cdots}}\,} = \frac{1}{\,3-\dfrac{12}{\,1-\dfrac{6}{3-\dfrac{20}{1-\cdots}}\,}\,}\end{align*}$$
and continue in this manner at each step to get the second form.
Finally, to show $\beta_n=n^2$, we define
$$m_k=\int_0^1 s^k\frac{1}{2}s^{-1/2}ds=\int_0^1 u^{2k}\,du=\frac{1}{2k+1}$$
and let
$$\Delta_n=\det\bigl(m_{i+j}\bigr)_{0\le i,j\le n}$$
with the convention $\Delta_{-1}:=1$ and $\Delta_0=m_0$. By writing the orthogonality linear systems for the polynomial coefficients and solving them by Cramer’s rule, one gets
$$\beta_n=\frac{\Delta_n\Delta_{n-2}}{\Delta_{n-1}^2}$$
for $n\ge 1$. On the other hand, it is also possible to show (using Induction or otherwise) that
$$\Delta_n =\frac{2^{\,n(n+1)}\left(\prod_{m=1}^n m!\right)^{2}}{\prod_{k=0}^{2n}(2k+1)^{\min(k+1,2n-k+1)}}$$
for $n\ge 1$. Plugging this into the equation of $\beta_n$ now completes the proof.
