Show that the series $\sum_{k=0}^{\infty}(-1)^k\frac{1}{2k+1}$ converges to $\pi/4$.
Here is my draft of a proof.
Abel's theorem states that suppose for a nondegenerate $[a,b]$. If $f(x):=\sum_{k=0}^{\infty}a_x(x-x_0)^k$ converges on $[a,b]$, then $f(x)$ is continuous and converges uniformly on $[a,b]$. Define $f(x)$ as $\frac{1}{1-x}$ and additionally as the power series centered at $x_0=0$ as $1+x+x^2+x^3+\cdots$. In other words $$f(x)=\frac{1}{1-x}:=\sum_{k=0}^{\infty}1\cdot(x)^k.$$ The radius of convergence can be found by letting $$r=\lim_{k\to\infty}|\frac{1\cdot(x)^{k+1}}{1\cdot(x)^k}|=|x|\lim_{k\to\infty}|1|=|x|.$$ By the ratio test, $r$ converges if $0\leq r<1$, or $|x|<1$ expressed by the interval $(-1,1)$.
It is now shown that $f(X)$ converges on a nondegenerate interval $(-1,1)$ and $f(x):=\sum_{k=0}^{\infty}a_x(x-x_0)^k$ where $a_k=1$ and $x_0=0$, therefore $f(x)$ converges uniformly on (-1,1) by Abel's theorem.
Next let $g(x)=f(-x^2)$. $g$ will still converge uniformly on $(-1,1)$ as $|-x^2|\leq|x|$ when $0\leq x<1$. As a result, $$g(x)=\frac{1}{1-(-x)^2}=\frac{1}{1+x^2}:=1-x^2+x^4-x^6+\cdots=\sum_{k=0}^{\infty}(-1)^k(x^{2k}).$$ Because $g(x)$ is uniformly convergent, it follows that $g$ is integrable on $(-1,1)$ and one can perform term-by-term integration on the series by $$\int g(x)dx=\int (1-x^2+x^4-x^6+\cdots)dx$$ from theorem 7.14. Here, $$G(x)=\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$
Evaluating $G(x)$ at 1 gives $$G(1)=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots=\sum_{k=0}^{\infty}(-1)^k\frac{1}{2k+1}=\arctan(1)=\frac{\pi}{4}$$
This is not entirely correct.
The series $\sum_{k=0}^\infty (-1)^kx^{2k}$ converges to $1/(1+x^2)$ on $(-1,1)$, but the convergence is not uniform. As a power series with convergence radius $1$, it does converge uniformly on any compact interval $[a,b] \subset (-1,1)$ -- a fact you cite but apply incorrectly. Consequently, you are premature in justifying termwise integration over $[0,1]$ on the basis of uniform convergence.
To see that the convergence is not uniform on the open interval, note that we have a geometric series where
$$\tag{*}\sum_{k=0}^n (-1)^kx^{2k} = \frac{1 - (-x^2)^{n+1}}{1+x^2},$$
and, as $n \to \infty$,
$$\sup_{x \in (-1,1)}\left|\sum_{k=0}^n (-1)^kx^{2k} - \frac{1 }{1+x^2}\right|= \sup_{x \in (-1,1)}\frac{(x^2)^{n+1}}{1+x^2} > \frac{((\sqrt{1-1/n})^2)^{n+1}}{1+(\sqrt{1-1/n})^2} \\ = \frac{(1-1/n)^{n+1}}{2-1/n} \to\frac{e^{-1}}{2} \neq 0$$
Solution 1
Using uniform convergence on the compact interval $[0,z] \subset (-1,1)$ we can integrate termwise to obtain
$$\sum_{k=0}^\infty \frac{(-1)^{k}z^{2k+1}}{2k+1} = \int_0^z \sum_{k=0}^\infty (-1)^{k}z^{2k} = \int_0^z \frac{dx}{1+x^2} = \arctan(z)$$
By Abel's limit theorem we have $$\sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} = \lim_{z \to 1-}\sum_{k=0}^\infty \frac{(-1)^{k}z^{2k+1}}{2k+1} = \lim_{z \to 1-}\arctan(z) = \frac{\pi}{4},$$
since the series on the LHS converges by the alternating series test.
Solution 2
A more direct way is to use (*) and apply termwise integration to the partial sum, which is always permissible, to obtain
$$\tag{**}\sum_{k=0}^n \frac{(-1)^{k}}{2k+1} - \frac{\pi}{4}= \int_0^1 \sum_{k=0}^n (-1)^kx^{2k} \, dx - \int_0^1 \frac{dx}{1+x^2} = (-1)^{n+1} \int_0^1 \frac{x^{2n+2}}{1+x^2}\,dx$$
Note that the RHS of (**) converges to $0$ as $n \to \infty$, by virtue of the estimate
$$\left|(-1)^{n+1} \int_0^1 \frac{x^{2n+2}}{1+x^2}\,dx\right| \leqslant \int_0^1 x^{2n+2} \, dx = \frac{1}{2n+3}$$
Therefore,
$$\lim_{n \to \infty}\sum_{k=0}^n \frac{(-1)^{k}}{2k+1} = \frac{\pi}{4}$$