Since the derivative of $\arctan (x)$ is $1/(x^2 + 1)$, by using the geometric series and
taking the integral term by term, we arrive at
$$
\arctan (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1} }{2n + 1}.
$$
I know that the radius of convergence is $R=1$.
This is then often used to give a series expression for $\pi$ by evaluating at $x=1$.
What I am confused with is, how do we know that at $x=1$, $\arctan (x)$ is the value of the series evaluated at $x=1$? Just because the Taylor series converges does not generally mean that it equals the function. Any clarification is appreciated! Thank you!
Best Answer
You are right. Just evaluating the series at $x=1$ is not valid and is a common mistake, but you may use Abel's theorem to make the argument rigorous.