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.

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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.
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.
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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