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?

The Taylor series expansion for the arctangent function (see this) is
$$\begin{align}
\arctan(x)
&=\sum_{n=0}^\infty \frac {(-1)^n}{2n+1}x^{2n+1}=x-\frac 13 x^3+\frac 15x^5-\frac 17 x^7+\cdots
\end{align}$$
Putting $x=1$ gives
$$\arctan(1)=\color{red}{\frac {\pi}4}=\sum_{n=0}^\infty \frac {(-1)^n}{2n+1}$$
