The function provided is:

$f(x) = \frac{1}{1 + x^2}, \quad |x| < 1$

To express this in terms of a geometric series, we recall the standard geometric series expansion for $\frac{1}{1 - z}$, which is valid for $|z| < 1$:

$\frac{1}{1 - z} = \sum_{n=0}^{\infty} z^n, \quad \text{for} \ |z| < 1.$

To apply this to the given function $f(x) = \frac{1}{1 + x^2}$, we rewrite it as:

$f(x) = \frac{1}{1 - (-x^2)}.$

Now we can use the geometric series expansion with $z = -x^2$:

$f(x) = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \quad \text{for} \ |x| < 1.$

Thus, the geometric series representation of the function is:

$f(x) = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \quad \text{for} \ |x| < 1.$

Would you like more details on how to use this expansion or any further questions?

Here are some related questions you might find useful:

1. What is the radius of convergence of a geometric series?
2. How can you find the sum of a geometric series?
3. How would this series change if the function were $f(x) = \frac{1}{1 - x^2}$?
4. Can we use the series expansion to approximate values of $f(x)$ for small $x$?
5. How does this series expansion relate to Taylor or Maclaurin series?

Tip: When dealing with series expansions, always ensure the range of convergence (in this case $|x| < 1$) is valid before applying it.