Let’s break this down step by step to solve the problem.

Problem:

We start with: $\frac{1}{1-x} = \sum_{n=0}^\infty x^n \quad \text{(converges for } -1 < x < 1 \text{)}.$

We need to rewrite the given expression: $\frac{x}{1-9x^4}$ as a power series and identify its interval of convergence.

Step 1: Rewriting the given expression

The denominator $1 - 9x^4$ matches the format $1 - u$, where $u = 9x^4$. Substitute this into the standard series: $\frac{1}{1-u} = \sum_{n=0}^\infty u^n \quad \text{(converges for } |u| < 1 \text{)}.$ Here: $\frac{1}{1-9x^4} = \sum_{n=0}^\infty (9x^4)^n = \sum_{n=0}^\infty 9^n x^{4n}.$

Now include the numerator $x$: $\frac{x}{1-9x^4} = x \cdot \sum_{n=0}^\infty 9^n x^{4n} = \sum_{n=0}^\infty 9^n x^{4n+1}.$

Thus, the power series is: $\boxed{\sum_{n=0}^\infty 9^n x^{4n+1}}$

Step 2: Interval of convergence

For convergence, we require $|9x^4| < 1$, which simplifies to: $|x|^4 < \frac{1}{9} \quad \Rightarrow \quad |x| < \frac{1}{3}.$

Hence, the series converges for $|x| < \frac{1}{3}$. To determine whether the endpoints $x = \pm \frac{1}{3}$ are included, we test the series at these points:

1. At $x = \frac{1}{3}$: $9^n (\frac{1}{3})^{4n+1} = \frac{1}{3} \cdot (\frac{1}{3^4})^n = \frac{1}{3} \cdot \frac{1}{3^{4n}}$.
   • The terms do not approach 0 as $n \to \infty$, so the series diverges.
2. At $x = -\frac{1}{3}$: Similar reasoning applies, as $(-\frac{1}{3})^{4n+1}$ results in divergence.

Thus, the endpoints are not included.

Final Answer:

1. Power series: $\boxed{\sum_{n=0}^\infty 9^n x^{4n+1}}$.
2. Interval of convergence: $\boxed{(-\frac{1}{3}, \frac{1}{3})}$.
   • Left end included: $\boxed{\text{N}}$.
   • Right end included: $\boxed{\text{N}}$.

Would you like a detailed explanation of any step, or have additional questions?

Related Questions:

1. What happens if the numerator in the series changes to $x^2$?
2. How is the interval of convergence affected by different powers of $x$ in the denominator?
3. What is the radius of convergence of this series?
4. Can the interval of convergence change if the numerator is omitted?
5. What would the series look like if expanded term by term?

Tip:

Always check convergence at boundary points separately to ensure the series does or does not include them.