<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] 
I thought it might be instructive to present an approach that relies on elementary analysis only .  To that end, we now proceed.

First, note that $\int_0^1 x^{2n}\,dx=\frac{1}{2n+1}$.  Therefore, we can write
$$\begin{align}
\sum_{n=0}^N \frac{(-1)^n}{2n+1}&=\sum_{n=0}^N(-1)^n\int_0^1x^{2n}\,dx\\\\
&=\int_0^1 \sum_{n=0}^\infty (-x^2)^n\,dx\\\\
&=\int_0^1 \frac{1-(-x^2)^{N+1}}{1+x^2}\,dx\\\\
&=\frac{\pi}{4}+(-1)^N \int_0^1 \frac{x^{2N+2}}{1+x^2}\,dx\tag 1
\end{align}$$
Noting that by integrating by parts the integral on the right-hand side of $(1)$, with $u=\frac{1}{1+x^2}$ and $v=\frac{x^{2N+3}}{2N+3}$, we see that 
$$\begin{align}
\left| \int_0^1 \frac{x^{2N+2}}{1+x^2}\,dx\right|&=\left|\frac{1}{2(2N+3)}+\frac{1}{2N+3}\int_0^1 \frac{2x}{(1+x^2)^2}x^{2N+3}\,dx\right|\\\\
&\le  \frac{5/2}{2N+3}\\\\
&\to 0\,\,\text{as}\,\,N\to \infty\tag 2
\end{align}$$
Using the result from $(2)$, we see that 
$$\lim_{N\to \infty}\sum_{n=0}^N \frac{(-1)^n}{2n+1}=\frac{\pi}{4}+\lim_{N\to \infty}\left((-1)^N \int_0^1 \frac{x^{2N+2}}{1+x^2}\,dx\right)=\frac{\pi}{4}$$
and hence we arrive at the coveted result
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$
as expected.

Alternatively, we can appeal to the Dominated Convergence Theorem.  Since $-1\le -x^2\le 0$, we see that $\left|\frac{1-(-x^2)^{N+1}}{1+x^2}\right|\le \frac{2}{1+x^2}$.  Since $\int_0^1 \frac{2}{1+x^2}\,dx<\infty$, then the Dominated Convergence Theorem guarantees that 
$$\begin{align}
\lim_{N\to \infty}\int_0^1 \frac{1-(-x^2)^{N+1}}{1+x^2}\,dx
&=\int_0^1\lim_{N\to \infty}\left( \frac{1-(-x^2)^{N+1}}{1+x^2}\right)\,dx\\\\
&=\int_0^1\frac{1}{1+x^2}\,dx\\\\
&=\frac{\pi}{4}
\end{align}$$ 
as was to be shown!
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