I wanted to find a proof of the Leibniz $\pi$ formula, knowing that $\arctan(1) = \pi / 4.$ As such, I simply needed to find the Maclaurin Series of $\arctan$ and its rate of convergence. I want to know if this solution is correct: $$(\arctan x)' = \frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum _{n=0} ^\infty (-x^2)^n = \sum _{n=0} ^\infty (-1)^n x^{2n}$$ So then $$\arctan x = \int \sum _{n=0} ^\infty (-1)^n x^{2n} = \sum _{n=0} ^\infty \int (-1)^n x^{2n} = \sum _ {n=0} ^\infty (-1)^n \frac{x^{2n+1}}{2n+1} $$
and its rate of convergence (I tried to find it using the ratio test, but I obtained that if $x < 1$ the series converges and if $x = 1$ the ratio is inconclusive. If I am correct, how can I prove that at $x=1$ the series converges?).
What about going the other way?
$\begin{align} 1-\dfrac13+\dfrac15-\dfrac17 \ldots &= \int_0^1 \left(1-x^2+x^4-x^6+x^8 \ldots \right) \mathrm dx \\ &= \int_0^1 \dfrac1{1+x^2} \mathrm dx \\ &= \left[ \tan^{-1} (x) \right] \bigg|_0^1 \\ &= \color{red}{\boxed{\color{blue}{\dfrac{\pi}{4}}}} \end{align}$