<s>[INST] 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?
 [/INST] We can write $1-\sin{x}=2\cos^2{(x/2+\pi/4)}$, and I think we can now see the problem: all the roots have to be double roots. They'll be at $(4n + 1)\pi/2$ and $-(4n+3)\pi/2$, as you suggest, but all the factors have to be squared:
$$ 1-\sin{x} = A \prod_{n=0}^{\infty} \left(1-\frac{2x}{(4n+1)\pi}\right)^2 \left(1+\frac{2x}{(4n+3)\pi}\right)^2, $$
and we have to worry about the value of $A$ to make sure we have the right scaling, about which the zeros tell us nothing. Of course, we actually know that $A=1$ by putting $x=0$.

Okay, so
$$ 1-\sin{x} = \prod_{n=0}^{\infty} \left(1-\frac{2x}{(4n+1)\pi}\right)^2 \left(1+\frac{2x}{(4n+3)\pi}\right)^2 = \prod_{k=0}^{\infty} \left(1-(-1)^k\frac{2x}{(2k+1)\pi}\right)^2. $$
Now let's get to the series. The easiest way to differentiate a product is to differentiate the logarithm, since this turns it into a sum:
$$ \log{(1-\sin{x})} = 2\sum_{k=0}^{\infty} \log{\left(1-(-1)^k\frac{2x}{(2k+1)\pi}\right)}. $$
Now differentiate:
$$ -\frac{\cos{x}}{1-\sin{x}} = -\frac{4}{\pi}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{4k+1} \frac{1}{1-(-1)^{k}2x/((2k+1)\pi)}  $$
Putting $x=0$ and multiplying everything by $-1$ gives
$$ 1 = \frac{4}{\pi} \sum_{k=0} \frac{(-1)^k}{4k+1}, $$
as required.
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