Question

Use the comparison test to determine whether the following series converge. $$ \sum_{n=1}^{\infty} \frac{n^{1.2}-1}{n^{2.3}+1} $$

Short answer

The series converges.

Step-by-step solution

Identify the Dominant Terms

In the series \( \sum_{n=1}^{\infty} \frac{n^{1.2}-1}{n^{2.3}+1} \), the dominant term in the numerator is \( n^{1.2} \), and in the denominator, it's \( n^{2.3} \). Therefore, the series behaves similarly to \( \frac{n^{1.2}}{n^{2.3}} = \frac{1}{n^{1.1}} \) for large \( n \).

Find a Suitable Comparison Series

Since the dominant term comparison\( \frac{1}{n^{1.1}} \) suggests a potential comparison series, consider the series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \).

Analyze the Comparison Series

The series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \) is a \( p \)-series with \( p = 1.1 \). Since \( p > 1 \), this \( p \)-series converges.

Apply the Comparison Test

By the Comparison Test, if \( \frac{n^{1.2}-1}{n^{2.3}+1} \leq \frac{C}{n^{1.1}} \) for a constant \( C \) and sufficiently large \( n \), and \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \) converges, then \( \sum_{n=1}^{\infty} \frac{n^{1.2}-1}{n^{2.3}+1} \) also converges. For large \( n \), \( \frac{n^{1.2}-1}{n^{2.3}+1} \approx \frac{1}{n^{1.1}} \), satisfying the condition of the comparison test.

Key concepts

Comparison Test

The comparison test is a valuable tool in determining the convergence or divergence of a series. It involves comparing the series in question with another series whose convergence behavior is already known, often called the comparison series.
For the series \( \sum_{n=1}^{\infty} \frac{n^{1.2}-1}{n^{2.3}+1} \), we use the comparison test by comparing it with the series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \).
Here's how it works:

• If the terms of our series \( a_n = \frac{n^{1.2}-1}{n^{2.3}+1} \) are smaller or equal to the terms of a convergent comparison series \( b_n = \frac{1}{n^{1.1}} \) for all large \( n \) and some constant \( C \), then our original series also converges.
• If the terms \( a_n \) are larger or equal to those of a divergent series, then our series diverges.

In this case, because \( b_n \) converges and \( a_n \leq C\cdot b_n \) for all sufficiently large \( n \), we conclude that the original series converges.

Dominant Terms

When analyzing series, it's important to focus on the dominant terms within the numerator and the denominator. They help us simplify the series by approximating its behavior for large \( n \). This approach allows for easier comparison with known convergent or divergent series.In the given series \( \sum_{n=1}^{\infty} \frac{n^{1.2}-1}{n^{2.3}+1} \):

• The dominant term in the numerator is \( n^{1.2} \).
• The dominant term in the denominator is \( n^{2.3} \).

This simplification leads us to consider the ratio \( \frac{n^{1.2}}{n^{2.3}} = \frac{1}{n^{1.1}} \). This is a crucial step because it indicates that for large \( n \), the series behaves similarly to \( \frac{1}{n^{1.1}} \). Recognizing dominant terms simplifies our analysis and helps in choosing a suitable comparison series.

P-Series

A \( p \)-series is a series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) where \( p \) is a constant. The convergence or divergence of a \( p \)-series depends on the value of \( p \):

• If \( p > 1 \), the \( p \)-series converges.
• If \( p \leq 1 \), the \( p \)-series diverges.

In our comparison step, we consider the series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \), which is a \( p \)-series with \( p = 1.1 \). Since \( p = 1.1 > 1 \), this series converges. This information allows us to leverage the comparison test successfully. Understanding \( p \)-series is crucial because many series in mathematics are structured in this way, making them easier to analyze for convergence.