Consider the section of graph of the function \[f(x) = \frac{1}{x^n},\ \ \ \ x \in \langle 0, 1],\ n > 0.\] This section of graph is revolved around the \(x\)-axis, resulting in a solid of revolution.

For which values of \(n\) does this solid of revolution have a *finite* volume?

**Clarification**:

If you are uncomfortable revolving a graph involving a vertical asymptote: consider the volume of the solid of revolution for \(x \in \langle \varepsilon, 1]\), and take the limit for \(\varepsilon\to 0\).

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