How to Use Comparison test for Convergence
- If \(0≤an≤bn\) for all \(n\) large enough and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
- If \(an≥bn\) and \(\sum b_n\) diverges, then \(\sum a_n\) also diverges.
The Comparison Test decides a series by bounding it against one whose behavior is known: if a series is term-by-term smaller than a convergent series it converges, and if it is larger than a divergent series it diverges. Below: the conditions, and converging and diverging worked examples.
Here are two examples demonstrating the Comparison Test:
1. Converging Series Example:
Series: \(\sum_{n=1}^{\infty} \frac{2}{n^3 + 1}\)
Comparison Series: \(\sum_{n=1}^{\infty} \frac{1}{n^3}\)
- The terms \(\frac{2}{n^3 + 1}\) are smaller than \(\frac{2}{n^3}\) for all \(n \geq 1\), and we know that \(\sum \frac{1}{n^3}\) converges (since it’s a p-series with \(p = 3\)).
- By the Comparison Test, since the comparison series converges, the original series also converges.
2. Diverging Series Example:
Series: \(\sum_{n=1}^{\infty} \frac{1}{n \ln(n)}\)
Comparison Series: \(\sum_{n=1}^{\infty} \frac{1}{n}\)
- The series \(\frac{1}{n \ln(n)}\) is larger than \(\frac{1}{n^2}\) for large \(n\), and we know that the harmonic series \(\sum \frac{1}{n}\) diverges.
- By the Comparison Test, since the comparison series diverges, the original series diverges as well.
Frequently Asked Questions
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