Problem 19 - Single Variable Calculus

Question

Use the Limit Comparison Test to determine whether the series is convergent or divergent. \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n^{3}-n-1}}\)

Step-by-Step Solution

Answer

The given series is \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n^{3}-n-1}}\). To determine its convergence behavior, we used the Limit Comparison Test with the comparison series \(\frac{1}{n^{\frac{3}{2}}}\). After finding the limit of the ratio of the two series as n goes to infinity, we got \(\lim_{n \to \infty} \frac{1}{\sqrt{1-\frac{n}{n^{3}}-\frac{1}{n^{3}}}} = 1\), which is a finite nonzero number. Since the comparison series converges (as it's a p-series with p = 3/2 > 1), the given series also converges.

Step 1: Set up the limit of the ratio of the series

First, we'll find the limit of the ratio of the given series and the comparison series as n goes to infinity. \( \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n^{3}-n-1}}}{\frac{1}{n^{\frac{3}{2}}}} \)

Step 2: Simplify the limit expression

Next, we will simplify the expression inside the limit: \( \lim_{n \to \infty} \frac{n^{\frac{3}{2}}}{\sqrt{n^{3}-n-1}} \)

Step 3: Evaluate the limit

To evaluate this limit, we will divide both the numerator and the denominator by the highest power term of n in the denominator. \( \lim_{n \to \infty} \frac{\frac{n^{\frac{3}{2}}}{n^{\frac{3}{2}}}}{\frac{\sqrt{n^{3}-n-1}}{n^{\frac{3}{2}}}} \) Simplify to get: \( \lim_{n \to \infty} \frac{1}{\frac{\sqrt{n^{3}-n-1}}{n^{\frac{3}{2}}}} \) Now, we have: \( \lim_{n \to \infty} \frac{1}{\sqrt{1-\frac{n}{n^{3}}-\frac{1}{n^{3}}}} \)

Step 4: Apply the Limit Comparison Test

Since the limit we found is a finite nonzero number (1), we can now apply the Limit Comparison Test. The comparison series we used is \(\frac{1}{n^{\frac{3}{2}}}\), which is a p-series with p = 3/2 > 1, meaning it converges. Therefore, the given series, \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n^{3}-n-1}}\), exhibits the same convergence behavior, making it convergent as well.

Key Concepts

Convergence of Series

Understanding the convergence of an infinite series is a fundamental aspect of calculus. This concept helps us ascertain whether the sum of an infinitely long list of numbers reaches a definite limit or not. To put it simply, a series \( \[ \sum_{n=1}^{\infty} a_n \] \) converges if, as we add more terms, their sum approaches a specific finite value. If not, it's said to diverge.

One way to determine convergence is using the Limit Comparison Test. This test compares a series of interest with a known 'comparison' series. If the ratio of the terms of the series and a convergent comparison series approaches a finite positive number, then both series will have the same behavior—either both converge or both diverge.

In the given exercise, we investigated the series \( \[ \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^{3}-n-1}} \] \). Upon using the Limit Comparison Test with the known p-series \( \[ \frac{1}{n^{\frac{3}{2}}} \] \), it was determined it converges. Hence, we conclude both series show convergent behavior.

P-Series

A p-series is an infinite series of the form \( \[ \sum_{n=1}^{\infty} \frac{1}{n^{p}} \] \), where \( p \) is a real number. The convergence of a p-series depends on the value of \( p \):

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

This characteristic makes p-series an excellent candidate for a comparison series. When using the Limit Comparison Test, comparing an unknown series to a p-series can decisively establish convergence or divergence.

In our step-by-step solution, the comparison series was a convergent p-series with \( p = \frac{3}{2} > 1 \). Such information is crucial to determining the behavior of the original series without having to sum an infinite number of terms.

Infinite Series

An infinite series is a sum of infinite terms, typically formulated as \( \[ \sum_{n=1}^{\infty} a_n \] \), where \( a_n \) represents the nth term in the sequence. The main idea is that as \( n \) grows without bounds, one tries to understand the behavior of the sum of these infinite terms—an insight into whether they add up to a finite value (converge) or not.

When dealing with infinite series, we encounter various types, such as geometric series, harmonic series, alternating series, and, as previously mentioned, p-series. Each type has its own set of rules for determining convergence, and the Limit Comparison Test is one of many tools available in calculus to help with this determination.

The exercise focused on a series that doesn't immediately appear to be a standard type. However, by using the Limit Comparison Test with a p-series, we were able to categorize it as convergent, which is a nifty workaround when direct evaluation isn't feasible.

Calculus

Calculus is the branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It’s divided mainly into two areas: differential calculus and integral calculus.

In the context of infinite series—part of integral calculus—we use calculus concepts to understand the behavior of series as they approach infinity. This helps us model and analyze real-world phenomena such as calculating areas under curves (known as the definite integral) and understanding growth patterns in sequences and series.

The exercise solution leverages calculus, specifically the tactics of limits and series convergence. The efficient use of the Limit Comparison Test embodies the analytical power of calculus, as it simplifies complex problems by relating them to more familiar and solvable terms or series. Thus, calculus is not just a mathematical discipline but a problem-solving toolkit suitable for a plethora of applications in science and engineering.