Question

Use the Limit Comparison Test to determine the convergence of the given series; state what series is used for comparison. $$\sum_{n=1}^{\infty} \frac{1}{n^{2}-3 n+5}$$

Short answer

The series \( \sum_{n=1}^{\infty} \frac{1}{n^{2}-3n+5} \) converges; compared with \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).

Step-by-step solution

Identify the series to compare with

To apply the Limit Comparison Test, we need a series that we can easily determine the convergence of, and that resembles the given series. For the series \( \sum_{n=1}^{\infty} \frac{1}{n^{2}-3n+5} \), observe that as \( n \to \infty \), the expression \( n^2-3n+5 \) behaves like \( n^2 \). Therefore, we'll compare it with the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).

Check convergence of the comparison series

The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a p-series with \( p = 2 > 1 \), which is known to converge.

Apply the Limit Comparison Test

According to the Limit Comparison Test, we calculate the following limit:\[L = \lim_{{n \to \infty}} \frac{a_n}{b_n} = \lim_{{n \to \infty}} \frac{\frac{1}{n^2-3n+5}}{\frac{1}{n^2}} = \lim_{{n \to \infty}} \frac{n^2}{n^2-3n+5}\]Simplifying further:\[L = \lim_{{n \to \infty}} \frac{1}{1-\frac{3}{n}+\frac{5}{n^2}}\]As \( n \to \infty \), \( \frac{3}{n} \to 0 \) and \( \frac{5}{n^2} \to 0 \), so:\[L = \frac{1}{1} = 1\]

Interpret the result of the Limit Comparison Test

Since \( L = 1 \) and \( 0 < L < \infty \), by the Limit Comparison Test, both series \( \sum_{n=1}^{\infty} \frac{1}{n^2-3n+5} \) and \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) either converge or diverge together. Since \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is known to converge, \( \sum_{n=1}^{\infty} \frac{1}{n^2-3n+5} \) also converges.

Key concepts

series convergence

Understanding the concept of series convergence is key to analyzing infinite series. A series, which is the sum of the terms of a sequence, converges if the sequence of its partial sums approaches a finite limit. If the partial sums continue to grow without bound, we say the series diverges. Convergence is central to many areas of mathematics because it allows for the approximation and evaluation of infinite processes. To determine convergence, we can use various tests, including the Limit Comparison Test utilized in our exercise. This helps compare the given series to another, more familiar series, to judge convergence behavior.

• Once a series is known to converge, we can infer that the infinite sum has a meaningful value.
• Convergence findings affect the practical usability of series in problem-solving scenarios, such as in physics and engineering.

p-series

A p-series is a specific type of series that takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The behavior of p-series is determined by the value of \( p \):

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

In our example, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is used as a comparison series. It's a well-known p-series with \( p = 2 \), which is greater than 1, ensuring its convergence. Understanding p-series is beneficial because it provides a foundation for testing other series' convergence using this known behavior.

limit calculation

Even though series convergence tests vary, many involve calculating limits to compare series terms directly. The Limit Comparison Test, for instance, involves evaluating the limit of the ratio of the terms from two series as \( n \to \infty \). In this particular problem, we computed:\[L = \lim_{{n \to \infty}} \frac{n^2}{n^2 - 3n + 5}\]Breaking it down simplifies to:\[L = \lim_{{n \to \infty}} \frac{1}{1 - \frac{3}{n} + \frac{5}{n^2}}\]As both \( \frac{3}{n} \) and \( \frac{5}{n^2} \) tend towards zero with increasing \( n \), the expression simplifies to \( L = 1 \). Calculating limits helps establish critical information about the behavior of the series, providing insight into whether two series share convergence properties.

comparison series

The choice of a suitable comparison series is pivotal in convergence testing. A comparison series should resemble the given series as \( n \to \infty \) and have a known convergence status. In our exercise, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2 - 3n + 5} \) is compared to \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). The logic is that when \( n \) grows large, the expression \( n^2 - 3n + 5 \) approximates \( n^2 \), making \( \frac{1}{n^2 - 3n + 5} \) akin to \( \frac{1}{n^2} \).

• A well-chosen comparison series simplifies the task of determining convergence by leveraging known results.
• It minimizes complex calculations while offering a robust framework for comparison.

This method is invaluable as it reduces endless calculations and directly leads to the conclusion about the given series' convergence.