Question 9.14: π = 4 ∑n=0^∞ (−1)^n/2n +1.

Using the Taylor series for the binomial function, we have, for all x ∈ (−1, 1),

\frac{1}{1 − x}= \sum\limits_{n=0}^{∞}{x^{n}},

⇒ \frac{1}{1 + x} = \frac{1}{1 − (−x)} = \sum\limits_{n=0}^{∞}{(−1)^{n}x^{n}},

⇒ \frac{1}{1 + x^{2}} = \sum\limits_{n=0}^{∞}{(−1)^{n}x^{2n}}.

Integrating both sides of the last equation, and using Theorem 9.8, we obtain

arctan u = \sum\limits_{n=0}^{∞}{\frac{(−1)^{n}u^{2n+1}}{2n + 1}},  u ∈ (0, 1).

Since the alternating series \sum{(−1)^{n}/(2n+1)} converges, we may apply Abel’s continuity theorem to obtain

\underset{u→1^{−}}{\lim} \arctan u = \sum\limits_{n=0}^{∞}{\frac{(−1)^{n}}{2n+1}},

noting that the left-hand side of this equation is π/4.