Employing the Limit Comparison Test to Analyze Series Convergence

The Limit Comparison Test decides whether a series converges by comparing it with a series whose behavior is already known. Take the limit of the ratio of their terms: if that limit is a positive finite number, both series do the same thing. Worked converging and diverging examples follow.

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}) Check out online math resources for more practice.

• 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 the 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.

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