Employing the Limit Comparison Test to Analyze Series Convergence

The Limit Comparison Test is a method for determining the convergence or divergence of a series by comparing it to another series with known behavior. Given two series \(\sum a_n\) and \(\sum b_n\), the test involves finding the limit of the ratio of their terms as \(n\) approaches infinity: \(L = \lim_{n \to \infty} \frac{a_n}{b_n}\)

• If \(L\) is a positive, finite number, then both series will either converge or diverge together.
• If \(L = 0\) and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
• If \(L = \infty\) and \(\sum b_n\) diverges, then \(\sum a_n\) also diverges.

The test is particularly useful when the terms of the series are complex or difficult to compare directly. It simplifies the process by relating the series to a known benchmark series, such as p-series, for which convergence or divergence is already understood.

Here are two examples demonstrating the Limit Comparison Test:

1. Converging Series Example:

Series: \(\sum_{n=1}^{\infty} \frac{2^n}{n^3}\)

Comparison Series: \(\sum_{n=1}^{\infty} \frac{1}{n^3}\)

Apply the Limit Comparison Test:

• The terms of the given series \(\frac{2^n}{n^3}\) grow faster than \(\frac{1}{n^3}\), but the exponential factor makes the series grow rapidly.
• Calculate the limit of the ratio: \(\lim_{n \to \infty} \frac{\frac{2^n}{n^3}}{\frac{1}{n^3}} = \lim_{n \to \infty} 2^n \) This limit approaches infinity, meaning the series diverges because \( \sum \frac{1}{n^3} \) converges and we know the series grows more quickly than the comparison.

2. Diverging Series Example:

Series: \(\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n}}\)

Comparison Series: \(\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}\)

Apply the Limit Comparison Test:

• The given series has terms similar to \(\frac{1}{n^{3/2}}\) for large \(n\).
• Calculate the limit of the ratio: \(\lim_{n \to \infty} \frac{\frac{1}{n \sqrt{n}}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{1}{1} = 1\)
• Since the limit is a positive finite number and \(\sum \frac{1}{n^{3/2}}\) converges (p-series with \(p = 3/2)\), the original series converges.