Is there an easy way of finding the taylor series for $1/(1+x^2)$? I was trying to calculate the fourth derivative and then I just gave up.
Recall that a geometric series can be represented as a sum by

$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty}x^n \quad \quad|x| &lt;1$$

Then we can simply manipulate our equation into that familiar format to get

$$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty}(-x^2)^{n} = \sum_{n=0}^{\infty}(-1)^n x^{2n}$$

Fun Alternative:

Note that $$\frac{d}{dx} \arctan x = \frac{1}{1+x^2}$$
and that the Taylor Series for $\arctan x$ is

$$\arctan x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$

$$\implies \frac{d}{dx} \arctan x = \frac{d}{dx}\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$

Interchange the sum and $\frac{d}{dx}$ (differentiate term-by-term) on the RHS to get

\begin{eqnarray*}
\frac{1}{1+x^2} &amp;=&amp; \sum_{n=0}^{\infty}\frac{d}{dx}(-1)^n\frac{x^{2n+1}}{2n+1}\\
&amp;=&amp; (-1)^n \frac{1}{2n+1}\sum_{n=0}^{\infty} \frac{d}{dx}x^{2n+1}\\
&amp;=&amp; (-1)^n \frac{1}{2n+1}\sum_{n=0}^{\infty} (2n+1)x^{2n}\\
&amp;=&amp; \sum_{n=0}^{\infty} (-1)^n\frac{(2n+1)x^{2n}}{2n+1}\\
\end{eqnarray*}

Cancel the $(2n+1)$ on the RHS to arrive at
$$\frac{1}{1+x^2}  = \sum_{n=0}^{\infty}(-1)^n x^{2n}$$
Yes. $1-x^2+x^4-x^6+x^8-x^{10} \cdots$