Problem 22 - Single Variable Calculus

Question

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

Step-by-Step Solution

Answer

Using the Limit Comparison Test and comparing the given series \(\sum_{n=2}^{\infty} \frac{\ln n}{n^{3}-1}\) with the convergent p-series \(\sum_{n=2}^{\infty} \frac{1}{n^2}\), we calculate the limit of the ratio of the terms as n approaches infinity. After simplifying the limit expression and applying L'Hôpital's Rule twice, we find that \(\lim_{n\to\infty} \frac{2\ln n + 2 + 1}{6n} = 0\). Since the limit is a finite positive value, both series have the same behavior. So, the given series is convergent.

Step 1: Identify the comparable series

We will compare the given series \(\sum_{n=2}^{\infty} \frac{\ln n}{n^{3}-1}\) with the convergent p-series \(\sum_{n=2}^{\infty} \frac{1}{n^2}\) (p=2, which is greater than 1).

Step 2: Set up the Limit Comparison Test

In order to apply the Limit Comparison Test, we need to calculate the limit of the ratio of the terms of both series as n approaches infinity: \(\lim_{n\to\infty} \frac{\frac{\ln n}{n^{3}-1}}{\frac{1}{n^2}}\)

Step 3: Simplify the limit expression

Simplify the limit expression by taking the reciprocal of the denominator: \(\lim_{n\to\infty} \frac{\ln n}{n^{3}-1} \cdot \frac{n^2}{1}\) This simplifies to: \(\lim_{n\to\infty} \frac{n^2\ln n}{n^3 - 1}\)

Step 4: Calculate the limit

To evaluate this limit, we can use L'Hôpital's Rule. Since both the numerator and denominator are growing without bound as n approaches infinity, we can take derivatives of the numerator and denominator with respect to n: Numerator derivative: \(\frac{d}{dn}(n^2\ln n) = 2n\ln n + n\) Denominator derivative: \(\frac{d}{dn}(n^3 - 1) = 3n^2\) Now, the limit becomes: \(\lim_{n\to\infty} \frac{2n\ln n + n}{3n^2}\)

Step 5: Apply L'Hôpital's Rule again

Since the limit still has both the numerator and denominator growing without bound as n approaches infinity, we apply L'Hôpital's Rule one more time: Numerator derivative: \(\frac{d}{dn}(2n\ln n + n) = 2\ln n + 2 + 1\) Denominator derivative: \(\frac{d}{dn}(3n^2) = 6n\) Now, the limit becomes: \(\lim_{n\to\infty} \frac{2\ln n + 2 + 1}{6n}\)

Step 6: Evaluate the final limit

As n approaches infinity, \(\ln n\) grows without bound, but the denominator, \(6n\), grows faster. Therefore, the limit is: \(\lim_{n\to\infty} \frac{2\ln n + 2 + 1}{6n} = 0\)

Step 7: Determine the behavior of the series

Since the limit of the ratio of the terms is 0, which is a finite positive value, by the Limit Comparison Test, the behavior of our given series \(\sum_{n=2}^{\infty} \frac{\ln n}{n^{3}-1}\) matches the behavior of the series \(\sum_{n=2}^{\infty} \frac{1}{n^2}\), which is convergent. Therefore, the given series is convergent.

Key Concepts

Convergent Series

A convergent series is a sequence of numbers that, when you add them up, the total sum approaches a finite limit.

Infinity is a tricky concept, but here’s the gist: if we keep adding terms of a sequence indefinitely and the total gets closer and closer to a certain number, we call this behavior convergence. Now, not all series are well-behaved. Some are rogue — their sum just keeps ballooning to infinity. Those are the divergent ones.

In the case of the series \(\sum_{n=2}^{\infty} \frac{\ln n}{n^{3}-1}\), we suspect it might approach a finite limit. To prove it, we use methods like the Limit Comparison Test, which sidekicks with a known convergent series, helping us infer the behavior of our series of interest.

L'Hôpital's Rule

Picture this: you’re trying to find the limit of a fraction, but every time you plug in the number you're approaching, the world falls apart — you get 0 divided by 0 or infinity over infinity. Panic? Not if you remember L'Hôpital's Rule! It says you can take that pesky fraction and replace it with a new one, made by taking derivatives of the top and bottom.

If the pesky limit still refuses to be a specific number and gives you a form like 0/0 or ∞/∞, just apply L'Hôpital's Rule again. Keep doing it until you get a straightforward answer, or you realize the original limit was indeed an abyss of mathematical despair after all (in which case, time to explore other techniques!).

In our textbook solution, we’ve tamed the wild \(\lim_{n\to\infty} \frac{n^{2}\ln n}{n^{3} - 1}\) down to zero by using L'Hôpital's Rule—twice. Patience and derivatives, that's the secret sauce.

P-series

The p-series is a family of series given by \(\sum_{n=1}^\infty \frac{1}{n^p}\), where ‘p’ is a positive real number. This family has clear favorites: only the ones with ‘p’ greater than 1 are convergent, meaning their sums cozy up to a finite limit. If p is 1 or less, sorry, the sum heads off to infinity; not quite ideal if you're looking for convergence.

The wonderful thing about p-series is their predictability in the convergence department, making them a perfect benchmark to compare to other more unpredictable series, like our \(\sum_{n=2}^{\infty} \frac{\ln n}{n^{3}-1}\) series. We chose \(\sum_{n=2}^\infty \frac{1}{n^2}\) (a convergent p-series with p=2) as a straightforward comparison point.

Infinite Series Calculus

In calculus, when we're confronting an infinite series, we're met with the challenge of determining its behavior: will it converge to a snug little finite sum, or will it be a rebel without a cause and diverge to infinity?

To make this decision, we have a toolkit of tests at our disposal: the Limit Comparison Test, the Ratio Test, the Integral Test, and a few others, each with its own savvy way of sniffing out convergence or divergence.

Applying these tools requires a mix of intuition and technique. Our textbook series is a testament to this dance of analytical moves: a clear strategy, understanding of series behavior, and the precise application of tools like L'Hôpital’s Rule and the Limit Comparison Test. With them, we've shown that just like its p-series counterpart, our series will indeed converge.