Question: User: $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots$ for $|x| < 1$.
1. **State the problem:** We want to understand the infinite series expansion of the function $$\frac{1}{1-x}$$ for values of $x$ such that $|x| < 1$.
2. **Formula used:** The formula for the sum of a geometric series is $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$ when $|x| < 1$.
3. **Explanation:** This means that the function $$\frac{1}{1-x}$$ can be expressed as an infinite sum of powers of $x$ starting from $x^0 = 1$.
4. **Intermediate work:** Writing out the first few terms explicitly:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots$$
5. **Important rule:** The series converges only if the absolute value of $x$ is less than 1, i.e., $|x| < 1$.
6. **Summary:** The infinite series expansion of $$\frac{1}{1-x}$$ is $$1 + x + x^2 + x^3 + \dots$$ valid for $|x| < 1$.