Power Series Expansion for Real Arctangent Function

Theorem
The arctangent function has a Taylor series expansion:


 * $\arctan x = \begin {cases} \ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1} & : -1 \le x \le 1 \\

\ds \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } & : x \ge 1 \\ \ds -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } & : x \le -1 \end {cases}$

That is:


 * $\arctan x = \begin {cases} x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 - \cdots & : -1 \le x \le 1 \\

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

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

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 $\sequence {\dfrac {x^{2 n + 1}} {2 n + 1} }$ 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 $x \le -1$:

Also see

 * Power Series Expansion for Real Arcsine Function
 * Power Series Expansion for Real Arccosine Function
 * Power Series Expansion for Real Arccotangent Function
 * Power Series Expansion for Real Arcsecant Function
 * Power Series Expansion for Real Arccosecant Function