Power Series Expansion for Real Arctangent Function

Theorem
The arctangent function has a Taylor series expansion:


 * $\arctan x = \begin{cases} \displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {2 n + 1} & : -1 \le x \le 1

\\ \displaystyle \frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac 1 {\left({2 n + 1}\right) x^{2 n + 1} } & : x \ge 1 \\ \displaystyle -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac 1 {\left({2 n + 1}\right) x^{2 n + 1} } & : x \le -1 \end{cases}$

That is:


 * $\arctan x = \begin{cases} \displaystyle x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \cdots & : -1 \le x \le 1

\\ \displaystyle \frac \pi 2 - \frac 1 x + \frac 1 {3 x^3} - \frac 1 {5 x^5} + \cdots & : x \ge 1 \\ \displaystyle -\frac \pi 2 - \frac 1 x + \frac 1 {3 x^3} - \frac 1 {5 x^5} + \cdots & : x \le -1 \end{cases}$

Proof
From Sum of Infinite Geometric Progression:
 * $(1): \quad \displaystyle \sum_{n \mathop = 0}^\infty \left({-x^2}\right)^n = \frac 1 {1 + x^2}$

for $-1 < x < 1$.

From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:

For $-1 \le x \le 1$, the sequence $\left \langle{\dfrac {x^{2 n + 1}} { 2n + 1} }\right \rangle $ is decreasing and converges to zero.

Therefore the series converges in the given range by the Alternating Series Test.

Now consider the case $x \ge 1$:

We also have:

Substituting $x$ for $-x$ gives us the expansion for $x \le -1$: