To find the power series representation of the function $f(x) = \frac{5}{1 - x^6}$ and determine its interval of convergence, let's proceed step-by-step.

Step 1: Rewrite the Function

We can express $f(x)$ in a form that resembles a geometric series. Notice that: $f(x) = \frac{5}{1 - x^6}$ This can be written as: $f(x) = 5 \cdot \frac{1}{1 - x^6}$

Step 2: Apply the Geometric Series Expansion

The standard geometric series expansion is: $\frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n \quad \text{for} \quad |u| < 1$ Here, we set $u = x^6$. Then: $\frac{1}{1 - x^6} = \sum_{n=0}^{\infty} (x^6)^n = \sum_{n=0}^{\infty} x^{6n}$ Thus, we can write: $f(x) = 5 \cdot \sum_{n=0}^{\infty} x^{6n} = \sum_{n=0}^{\infty} 5 x^{6n}$

Step 3: Determine the Interval of Convergence

The power series $\sum_{n=0}^{\infty} x^{6n}$ converges when $|x^6| < 1$, which simplifies to: $|x| < 1$ Therefore, the interval of convergence for this series is: $|x| < 1$ or in interval notation: $(-1, 1)$

Final Answer

The power series representation of $f(x)$ is: $f(x) = \sum_{n=0}^{\infty} 5 x^{6n}$ with an interval of convergence $|x| < 1$ (or $(-1, 1)$).

Would you like more details on any part of this solution?

Here are some related questions:

1. How would you find the radius of convergence of a general power series?
2. Can this approach be extended to functions with different exponents in the denominator?
3. How does the geometric series formula help in finding power series representations?
4. What happens to the power series if the function had a different constant, say $\frac{10}{1 - x^6}$?
5. How does the interval of convergence change if the exponent inside is $x^4$ instead of $x^6$?

Tip: When working with power series, always verify the form against the geometric series for easy expansion.