$\pi$ as an Infinite Series using Taylor Expansion on Equation of a Circle If we write $\ y = \sqrt{r^2-x^2}$ for the equation of a circle in Cartesian coordinates with radius $r$ and perform a Taylor expansion on this equation, then integrate term by term from $0$ to $r$, giving the area of a quarter of a circle. Then, isolating $\pi$, all $r$ terms conveniently drop out and the expression becomes:
$$\pi = 4 - \frac{4}{3!}- \frac{12}{5!}- \frac{180}{7!}- \frac{6300}{9!}- \frac{396900}{11!} - ...$$ 
This can be written as:
$$\ \pi = \sum_{n=1}^\infty \frac{-4[(2n-3)!!]^2}{(2n-3)(2n-1)!}$$
I attach here a WolframAlpha Link of this evaluated to 2700 terms.
I haven't seen this expression anywhere else. I was wondering if you have, or if you know if this is valid?
 A: You can do the same thing using a unit circle (this is basic dimensionality analysis).
\begin{align}\frac{\pi}{4} &= \int_0^1 \sqrt{1-x^2} \,dx
= \int_0^1 \left((1+(-x^2)\right)^{1 /2}\, dx\\
&= \int_0^1 \sum_{k=0}^{\infty}\binom{1 /2}{k} (-x^2)^k\, dx
\overset{(*)}{=} \sum_{k=0}^{\infty} (-1)^k\binom{1 /2}{k}\int_0^1 x^{2k}\, dx\\
&= \sum_{k=0}^{\infty} (-1)^k \frac{\binom{1 /2}{k}}{2k+1}
\end{align}
Now you can show that the mess of double factorials simplifies to
$$\binom{1/ 2}{k} = -\frac{1}{2k-1}\binom{2k}{k}\left(-\frac{1}{4}\right)^k,\quad\text{for $k> 0$, and}\quad \binom{1/ 2}{0} = 1$$
hence
$$\sum_{k=0}^{\infty} (-1)^k \frac{\binom{1 /2}{k}}{2k+1}
= 1 - \sum_{k=1}^{\infty} \frac{\binom{2k}{k}}{(2k-1)(2k+1)4^k}
= 1 - \sum_{k=1}^{\infty} \frac{\binom{2k}{k}}{(4k^2-1)4^k}$$
So we get a form that is similar to yours:
$$
\pi = 4 - \sum_{k=1}^{\infty} \frac{\binom{2k}{k}}{(4k^2-1)4^{k-1}}
$$
The central binomial coefficients $\binom{2k}{k}$ satisfy:
$$\binom{2k}{k} \leq \frac{4^k}{\sqrt{3k+1}},\quad\text{for $k\geq 1$}$$
which may be proven by induction. Then
\begin{align}
\sum_{k=1}^{\infty} \frac{\binom{2k}{k}}{(4k^2-1)4^{k-1}}
&\leq\sum_{k=1}^{\infty} \frac{4^k}{\sqrt{3k+1}(4k^2-1)4^{k-1}}
= \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{\sqrt{3k+1}(2k-1)(2k+1)}\\
&\leq \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{\sqrt{3k+1}(2k-1)(2k+1)}
\leq \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{\sqrt{2k-1}(2k-1)(2k-1)}\\
&\leq \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{(2k-1)^{5/2}}
\leq \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^{5/2}}
< \infty
\end{align}
with the latter convergence given by the $p$-series test.
As the partial series are positively increasing and bounded above, they must converge. This proves the convergence for the expression we gave.
Finally I somewhat skipped over the $(*)$ way back at the beginning of this answer, where we assume that we can swap the integral and summation. We eventually showed the terms are non-negative and converge, hence by Tonelli's Theorem we were justified in swapping the order of integrals (and sums).
