2+(-1)* k2 3. 10.5.27 Note that 2+ (-1)k < 3 for k > 1. It follows that 3 for k > 1. Because k2 k=1 1 E and 3. converges. It follows that the given series converges by 3. k2 k2 is a convergent p-series, k2 k=1 k=1 k=1 the Comparison Test.
2+(-1)* k2 3. 10.5.27 Note that 2+ (-1)k < 3 for k > 1. It follows that 3 for k > 1. Because k2 k=1 1 E and 3. converges. It follows that the given series converges by 3. k2 k2 is a convergent p-series, k2 k=1 k=1 k=1 the Comparison Test.
Chapter9: Sequences, Probability And Counting Theory
Section9.4: Series And Their Notations
Problem 55SE: The sum of an infinite geometric series is five times the value of the first term. What is the...
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For the part circled in green, can you show where the 3 came from?
![The image shows a mathematical expression labeled as problem number 27. The expression is an infinite series given by:
\[
\sum_{k=1}^{\infty} \frac{2 + (-1)^k}{k^2}
\]
Here, the series is the sum of terms starting from \(k = 1\) to infinity. Each term in the series is of the form \(\frac{2 + (-1)^k}{k^2}\).
Details:
- The numerator alternates between 3 and 1 due to the \((-1)^k\) term, which results in 3 when \(k\) is even and 1 when \(k\) is odd.
- The denominator is the square of \(k\), which increases with increasing \(k\).
This series explores the convergence of alternating terms adjusted for the squares of their indices.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F94f587c8-177e-4042-9792-23ea87be93b0%2Fj7hmfqv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image shows a mathematical expression labeled as problem number 27. The expression is an infinite series given by:
\[
\sum_{k=1}^{\infty} \frac{2 + (-1)^k}{k^2}
\]
Here, the series is the sum of terms starting from \(k = 1\) to infinity. Each term in the series is of the form \(\frac{2 + (-1)^k}{k^2}\).
Details:
- The numerator alternates between 3 and 1 due to the \((-1)^k\) term, which results in 3 when \(k\) is even and 1 when \(k\) is odd.
- The denominator is the square of \(k\), which increases with increasing \(k\).
This series explores the convergence of alternating terms adjusted for the squares of their indices.
![### Example 10.5.27
**Problem Statement:**
Note that \( 2 + (-1)^k \leq 3 \) for \( k \geq 1 \).
**Solution Steps:**
1. It follows that:
\[
\frac{2 + (-1)^k}{k^2} \leq \frac{3}{k^2} \quad \text{for} \quad k \geq 1.
\]
2. Because:
\[
\sum_{k=1}^{\infty} \frac{3}{k^2}
\]
is a convergent p-series.
3. The series:
\[
\sum_{k=1}^{\infty} \frac{1}{k^2}
\]
is a convergent p-series.
4. Therefore, the series:
\[
\sum_{k=1}^{\infty} 3 \left(\frac{1}{k^2}\right)
\]
converges.
5. It follows that the given series converges by the Comparison Test.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F94f587c8-177e-4042-9792-23ea87be93b0%2F5n6v05_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example 10.5.27
**Problem Statement:**
Note that \( 2 + (-1)^k \leq 3 \) for \( k \geq 1 \).
**Solution Steps:**
1. It follows that:
\[
\frac{2 + (-1)^k}{k^2} \leq \frac{3}{k^2} \quad \text{for} \quad k \geq 1.
\]
2. Because:
\[
\sum_{k=1}^{\infty} \frac{3}{k^2}
\]
is a convergent p-series.
3. The series:
\[
\sum_{k=1}^{\infty} \frac{1}{k^2}
\]
is a convergent p-series.
4. Therefore, the series:
\[
\sum_{k=1}^{\infty} 3 \left(\frac{1}{k^2}\right)
\]
converges.
5. It follows that the given series converges by the Comparison Test.
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