What comparison series would you use with the Comparison Test to determine whether
∑ (k = 1 to ∞) 1 / (k² + 1) converges?

Verified step by step guidance

1. Identify the given series: \( \sum_{k=1}^{\infty} \frac{1}{k^2 + 1} \). We want to determine if it converges using the Comparison Test.
2. Recall that for large values of \( k \), the term \( \frac{1}{k^2 + 1} \) behaves similarly to \( \frac{1}{k^2} \) because the \( +1 \) becomes insignificant compared to \( k^2 \).
3. Choose a comparison series that is simpler but related, such as \( \sum_{k=1}^{\infty} \frac{1}{k^2} \), which is a p-series with \( p = 2 > 1 \) and is known to converge.
4. Use the Comparison Test by noting that \( \frac{1}{k^2 + 1} < \frac{1}{k^2} \) for all \( k \geq 1 \), so if the larger series \( \sum \frac{1}{k^2} \) converges, then the original series also converges.
5. Conclude that the appropriate comparison series to use is \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) because it helps establish convergence of the original series by the Comparison Test.

Key Concepts

Comparison Test

The Comparison Test determines the convergence or divergence of a series by comparing it to another series with known behavior. If the terms of the given series are smaller than those of a convergent series, it also converges. Conversely, if the terms are larger than those of a divergent series, it diverges.

Choice of Comparison Series

To apply the Comparison Test effectively, select a simpler series whose convergence is known and whose terms closely resemble the original series. For ∑ 1/(k² + 1), the series ∑ 1/k² is a natural choice because for large k, 1/(k² + 1) behaves like 1/k².

Convergence of p-Series

A p-series ∑ 1/k^p converges if and only if p > 1. Since ∑ 1/k² is a p-series with p = 2, it converges. This fact helps conclude that ∑ 1/(k² + 1) also converges by comparison.