Question
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
Ask by Wells Tyler. in the United States
Mar 15,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The sequence \( \{r^n\} \) converges for \( -1 < r \leq 1 \) and diverges otherwise.
Solution
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.
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Mind Expander
The sequence \( \{r^n\} \) converges when \( |r| < 1 \). In this scenario, as \( n \) approaches infinity, the terms \( r^n \) get closer and closer to 0. Conversely, for \( |r| \geq 1 \), the sequence diverges—either growing indefinitely or oscillating without settling on a single value. So, the correct choice is B. The sequence \( \{r^{n}\} \) converges for \( -1 < r < 1 \), and diverges otherwise.