The given image shows the function $f(x)$ defined as a series and its equivalent closed-form expression:

$f(x) = \sum_{n \geq 0} x^n = \frac{1}{1 - x}$

This is a geometric series where the sum is taken from $n = 0$ to infinity. Here’s a detailed breakdown:

Geometric Series

A geometric series is a series of the form:

$\sum_{n=0}^{\infty} ar^n$

where $a$ is the first term and $r$ is the common ratio. For convergence of this series, $|r| < 1$ must hold.

Given Series

In your function:

$f(x) = \sum_{n=0}^{\geq 0} x^n$

Here, $a = 1$ and $r = x$. The sum of a geometric series is given by:

$\sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r}$

Applying it to the given series, we get:

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

Convergence

The series converges for $|x| < 1$, meaning:

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

Summary

The function $f(x)$ defined as $\sum_{n=0}^{\infty} x^n$ is a geometric series and can be expressed in closed form as $\frac{1}{1 - x}$ for $|x| < 1$.

Would you like more details or have any questions about this?

Here are 5 related questions to expand on this topic:

1. How can we derive the sum of a geometric series?
2. What are the conditions for the convergence of a geometric series?
3. Can you find the sum of the series $\sum_{n=0}^{\infty} 2^n x^n$?
4. How does the series behave when $x$ is outside the interval $|x| < 1$?
5. Can this series representation be used for other values of $x$ using analytic continuation?

Tip: When dealing with series, always check for the interval of convergence to ensure the series sum is valid.