Question

Use the Limit Comparison Test to determine the convergence of the given series; state what series is used for comparison. $$\sum_{n=1}^{\infty} \frac{n+5}{n^{3}-5}$$

Short answer

Convergent, compared to \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).

Step-by-step solution

Identify the Type of Series

The given series is \( \sum_{n=1}^{\infty} \frac{n+5}{n^{3}-5} \). This is a series where the terms appear to behave like a rational function similar to the form \( \frac{1}{n^2} \) or \( \frac{1}{n^p} \).

Propose a Comparison Series

For limit comparison test, we choose a series that has a similar form and is known to converge or diverge. Comparing with \( \frac{n}{n^3} = \frac{1}{n^2} \). The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a p-series with \( p = 2 \), which converges.

Apply the Limit Comparison Test

We compute the limit \( L \) of the ratio of the two sequences:\[ L = \lim_{n\to\infty} \frac{\frac{n+5}{n^3-5}}{\frac{1}{n^2}} = \lim_{n\to\infty} \frac{(n+5)n^2}{n^3-5} \]Simplifying gives:\[ L = \lim_{n\to\infty} \frac{n^3 + 5n^2}{n^3-5} \]Divide numerator and denominator by \( n^3 \):\[ L = \lim_{n\to\infty} \frac{1 + \frac{5}{n}}{1 - \frac{5}{n^3}} = 1 \]

Determine Convergence Based on the Limit

Since \( L = 1 \) and \( 0 < L < \infty \), by the Limit Comparison Test, the original series \( \sum_{n=1}^{\infty} \frac{n+5}{n^3-5} \) converges similarly to \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which we know converges.

Key concepts

series convergence

In mathematical analysis, when dealing with infinite series, determining whether a series converges or diverges is a central question. Convergence means that as you add infinitely many terms of the series, you approach a finite value. Conversely, divergence means that the sum grows without bound.
For the series \(\sum_{n=1}^{\infty} a_n\), we assess convergence by various tests such as the Limit Comparison Test, Ratio Test, or Integral Test, among others.

• Absolute Convergence: Occurs if \(\sum_{n=1}^{\infty} |a_n|\) converges. It implies the original series \(\sum_{n=1}^{\infty} a_n\) also converges.
• Conditional Convergence: The series \(\sum_{n=1}^{\infty} a_n\) converges, but \(\sum_{n=1}^{\infty} |a_n|\) does not.
• Divergence: If the terms do not approach zero, the series diverges. Also, if no convergence criterion is satisfied, the series diverges.

The Limit Comparison Test, like in our problem, is particularly useful for series that resemble a function whose convergence properties are already known, typically a p-series.

p-series

The p-series is a special type of series with a specific form that has well-known convergence properties. A p-series is represented as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) where \( p \) is a constant exponent. Understanding when a p-series converges or diverges is crucial in comparison tests.Convergence Criteria for p-series:

• If \( p > 1 \), the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges.
• If \( 0 < p \leq 1 \), the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) diverges.

In our current task, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is chosen as a comparison basis. This is a p-series with \( p = 2 \), which means it converges. Using this comparison helps in determining the behavior of series that have similar high-power terms and can be compared using the Limit Comparison Test.

rational functions

Rational functions play a key role in series and convergence tests because many terms in series are expressed as functions. A rational function is the ratio of two polynomials, like \( \frac{n+5}{n^3-5} \) in this exercise.

Key Characteristics of Rational Functions:

• Degree: The degree of a rational function affects its limit properties as \( n \to \infty \). If the degree of the numerator is less than the degree of the denominator, the function approaches zero.
• Comparison: Rational functions can simplify the process of using convergence tests like the Limit Comparison Test, due to the ease of analyzing asymptotic behavior.

In our example, the given series' term \( \frac{n+5}{n^3-5} \) behaves similarly to \( \frac{1}{n^2} \), a known-converging series, because the highest power in the denominator \( n^3 \) predominates, leading to a degree comparison with known series forms. This ability to simplify a term to a p-series form is indispensable when determining series convergence.