Why is the value of the series representation for $\arctan$ at $ x = 1$ necessarily $\arctan(1)$ Exercise 6.6.1. The derivation in Example 6.6.1 shows the Taylor series for $\arctan(x)$ is valid for all $x \in (−1,1)$. Notice, however, that the series also converges when $x = 1$. Assuming that $\arctan(x)$ is continuous, explain why the value of the series at $x = 1$ must necessarily be $\arctan(1)$. What interesting identity do we get in this case?
This is a question from Abbott's Understanding Analysis. My question is what exactly constitutes a proof that the series equals the function at a point.

Proof. Since $arctan(x)$ is continuous, for all $\epsilon > 0 $ there exists some $\delta > 0 $ such that $|1-x| < \delta $ implies that $|arctan(1) - arctan(x)| < \epsilon $. We also know that $$arctan(x) = x - \frac 1 3 x^3 + \frac 1 5 x^5 - \frac 1 7 x^7 + \cdots $$ when $ x \in (-1, 1)$. Hence  $|1-x| < \delta $ implies that $$|arctan(1) - ( x - \frac 1 3 x^3 + \frac 1 5 x^5 - \frac 1 7 x^7 + \cdots )| < \epsilon $$ for all $\epsilon>0$. Hence $$arctan(1) = 1 - \frac 1 3 1^3 + \frac 1 5 1^5 - \frac 1 7 1^7 + \cdots$$ as desired.
