For what values of \( r \) does the sequence \( \left\{r^{n}\right\} \) converge? Diverge? Select the correct choice below and fill in any answer boxes in your choice. A. The sequence \( \left\{r^{n}\right\} \) converges for \( r>\square \), and diverges otherwise. B. The sequence \( \left\{r^{n}\right\} \) converges for \( \square \leq r \leq \square \), and diverges otherwise. D. The sequence \( \left\{r^{n}\right\} \) converges for \( r \leq \square \), and diverges otherwise. E. The sequence \( \left\{r^{n}\right\} \) converges for \( \square

Ask by Wells Tyler. in the United States

Mar 15,2025

We wish to determine for which real numbers \( r \) the sequence \[ \{r^n\}_{n=1}^{\infty} \] converges and for which it diverges. **Step 1. Consider when \(|r| < 1\).** If \(|r| < 1\) then as \( n \to \infty \), the sequence satisfies \[ r^n \to 0. \] Thus, the sequence converges (to 0). **Step 2. Consider when \( r = 1\).** In this case, every term in the sequence is \[ 1^n = 1. \] So the sequence converges (to 1). **Step 3. Consider when \( r = -1\).** Then the sequence alternates: \[ -1, \, 1, \, -1, \, 1, \, \dots \] Since the sequence does not approach a single number, it diverges. **Step 4. Consider when \(|r| > 1\).** If \(|r| > 1\) then as \( n \to \infty \), the terms \( r^n \) become arbitrarily large in absolute value (and if \( r \) is negative, they also alternate in sign). In either case, the sequence diverges. **Conclusion** The sequence converges if and only if \[ -1 < r \leq 1. \] Among the provided choices, the one that corresponds to this is: E. The sequence \( \{r^n\} \) converges for \( \boxed{-1} < r \leq \boxed{1} \), and diverges otherwise.