Exercise 6.6.1

Question

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?

Ulisse Mini · Accepted Answer

Abel's theorem (Theorem 6.5.4) implies the series converges uniformly on $[0, 1]$.
  Combined with (Theorem 6.2.6) we see function the series converges to must be continuous.
  Taking limits shows this value must be $\arctan(1)$ giving the identity
  $$
  \arctan(1) = \frac{\pi}{4} = 1 - \frac 13 + \frac 15 - \frac 17 + \dots
  $$

Exercise 6.6.1

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Exercise 6.6.1

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