Question

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

Short answer

The series diverges by comparison to the harmonic series \( \sum_{n=2}^{\infty} \frac{1}{n} \).

Step-by-step solution

Identify the Given Series

We are given the series \( \sum_{n=2}^{\infty} \frac{n}{n^{2}-1} \). The task is to determine its convergence or divergence.

Simplify the Term

First, rewrite the term \( \frac{n}{n^2-1} \) as \( \frac{n}{(n-1)(n+1)} \). This simplifies our understanding of the comparison we need to make.

Choose a Comparison Series

Considering the term \( \frac{n}{n^2-1} \approx \frac{n}{n^2} = \frac{1}{n} \), choose the harmonic series \( \sum_{n=2}^{\infty} \frac{1}{n} \) for comparison, which is known to diverge.

Apply the Direct Comparison Test

Evaluate whether \( 0 < \frac{n}{n^2-1} \leq \frac{1}{n} \) for \( n \geq 2 \). Simplify to check if \( \frac{n}{n(n-1)} \leq \frac{1}{n} \), or equivalently, if \( \frac{1}{n-1} \leq \frac{1}{n} \). Since \( n-1 < n \), this inequality holds.

Conclude Using the Direct Comparison Test

Since \( \frac{n}{n^2-1} \leq \frac{1}{n} \) and \( \sum_{n=2}^{\infty} \frac{1}{n} \) diverges, the original series \( \sum_{n=2}^{\infty} \frac{n}{n^2-1} \) also diverges by the Direct Comparison Test.

Key concepts

Series Convergence

When studying series convergence, we are looking at whether the sum of an infinite series adds up to a finite number, or if it grows without bound. Convergence means the series settles on a number as you keep adding more terms, while divergence means it does not settle.
Understanding convergence is essential in calculus as it helps us analyze series involving integers and functions. In simpler terms, if we say a series is convergent, it stabilizes as we progress, whereas for divergent series, it just keeps growing.
To determine convergence, we often use tests like the Direct Comparison Test. These tests are designed to tell us if the series converges or diverges without necessarily finding the sum.
Common strategies include:

• Identifying patterns within the terms of the series.
• Comparing with other known series to make conclusions about convergence.
• Using algebraic transformations to simplify comparison series.

Understanding these fundamentals can simplify the complexities often seen in calculus.

Harmonic Series

The harmonic series is one of the most famous examples in mathematics, given by \( \sum_{n=1}^{\infty} \frac{1}{n} \). It's a series that diverges, meaning that as you add more terms, the sum does not stabilize to a certain number; instead, it increases indefinitely.
The harmonic series is considered a baseline in comparison testing because of its simple form. Despite the terms \( \frac{1}{n} \) getting smaller, their sum swells to infinity.
Interestingly, this happens because each additional term adds a small yet non-negligible amount to the total.
The divergence of the harmonic series is an essential concept, especially when using the direct comparison test, as it sets a foundational example of how a sequence of decreasing terms can still lead to an infinite sum.
When addressing problems like our original exercise, knowing the properties of the harmonic series can aid significantly in comparing and concluding series behaviors.

Inequalities in Convergence Testing

Inequalities are crucial tools in convergence testing. When we use the Direct Comparison Test, inequalities help us determine which series one should compare with. The idea is to show that each term in our series is either larger or smaller than a comparable term in a series with known convergence behavior.
For the Direct Comparison Test, we consider two series:

• Our series of interest, \( \sum a_n \).
• A known comparable series, \( \sum b_n \).

The test works using these inequalities:

• If \( 0 \leq a_n \leq b_n \) for all n beyond a certain point and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( 0 \leq b_n \leq a_n \) for all n beyond a certain point and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

These inequality conditions form the backbone of many convergence conclusions, allowing us to deduce the behavior of complex series effectively by comparing them to simpler, well-understood ones. In our original problem, using the inequality \( \frac{n}{n^2-1} \leq \frac{1}{n} \) allowed us to determine divergence by using what we know about the harmonic series.