Power series of ln(1+x) and arctan x

1. For $|t|<1$, $\frac1{1+t}=\sum_{n\ge0}(-1)^nt^n$. Integrating from $0$ to $x$ (allowed inside the radius of convergence):
$$\ln(1+x)=\sum_{n\ge0}\frac{(-1)^nx^{n+1}}{n+1}=x-\frac{x^2}2+\frac{x^3}3-\cdots$$
Likewise $\frac1{1+t^2}=\sum(-1)^nt^{2n}$ gives
$$\arctan x=\sum_{k\ge0}\frac{(-1)^kx^{2k+1}}{2k+1}=x-\frac{x^3}3+\frac{x^5}5-\cdots$$
Both have radius of convergence $1$.

2. At $x=1$. The two series at $x=1$ converge by the alternating series test, and by Abel's theorem (if a power series converges at an endpoint of its interval, its sum is continuous up to that endpoint), their sums equal the left limits $\ln2$ and $\arctan1=\frac\pi4$:
$$\ln2=1-\frac12+\frac13-\cdots,\qquad\frac\pi4=1-\frac13+\frac15-\cdots$$

Remark. Leibniz's series for $\frac\pi4$ is beautiful but converges very slowly: after $1000$ terms the error is still about $\frac1{2000}$. Machin-type formulas use $\arctan\frac15$ and $\arctan\frac1{239}$ instead.