Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $ 
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$

$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which occur at $x = \frac{(4n+1)\pi}{2} $ and $\frac{-(4n+3) \pi}{2}$
Equating this to its Taylor series, we get
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x)(1 + \frac{2}{3\pi}x)(1 - \frac{2}{5\pi}x)(1 + \frac{2}{7\pi}x)... $$
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x(1-\frac13)+O(x^2))(1 - \frac{2}{\pi}x(\frac15-\frac17)+O(x^2))... $$
$$1 - x + O(x^3) = 1 - \frac{2}{\pi}x(1-\frac13+\frac15-\frac17+...)+O(x^2)$$
$$x - O(x^3) = \frac{2}{\pi}x \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}-O(x^2)$$
Equating $x$ terms,
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \frac{\pi}{2}$$
However, I seem to be off by a factor of $2$ (actual answer is $\frac{\pi}{4} $). Does anyone see where this went wrong?

This is the value of the Dirichlet beta function at the input $s=1$, which is defined as the series
$$\beta(s)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$$
There is an integral representation related with the Gamma Function and the Dirichlet beta function.
$$\beta(s)\Gamma(s)=\int_0^\infty \frac{x^{s-1}e^{-x}}{1+e^{-2x}}\, \mathrm dx$$
Sketch of the proof:\begin{align}\mathcal{I}=\int_0^\infty \frac{x^{s-1}e^{-x}}{1+e^{-2x}}\, \mathrm dx&=\int_0^\infty x^{s-1}e^{-x}\bigg(\sum_{n=0}^\infty(-e^{-2x})^k\bigg)\\&=\sum_{n=0}^\infty (-1)^k\int_0^\infty x^{s-1}e^{-(2k+1)x}\, \mathrm dx\\&=\sum_{n=0}^\infty \frac{(-1)^k}{(2k+1)^s}\Gamma(s)\\&=\beta(s)\Gamma(s)\end{align}
In particular, we would find
\begin{align}\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}&=\beta(1)\Gamma(1)\\&=\int_0^\infty \frac{e^{-x}}{1+e^{-2x}}\, \mathrm dx\\&=\int_0^\infty \frac{e^{x}}{e^{2x}+1}\ \mathrm dx\\&=\int_0^\infty\frac{1}{1+t^2}\, \mathrm dt\, \text{, via substituting $t=e^x$}\\&=\frac{\pi}{2}\end{align}
Hence proved.
