Problem 18 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{3}{n+\sqrt{n}}\end{equation}

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{3}{n+\sqrt{n}} \) diverges by the limit comparison test with the harmonic series.

Step 1: Consider the Comparison Test

We will compare the series \( \sum_{n=1}^{\infty} \frac{3}{n+\sqrt{n}} \) to a series we can easily determine its convergence. A reasonable comparison is \( \sum_{n=1}^{\infty} \frac{1}{n} \). Both converge or diverge based on their behavior as \( n \to \infty \).

Step 2: Simplification for Comparison

Observe that \( \frac{3}{n+\sqrt{n}} \approx \frac{3}{n} \) for large \( n \), because \( \sqrt{n} \) becomes much smaller relative to \( n \). Thus, we can compare \( \frac{3}{n+\sqrt{n}} \) with \( \frac{3}{n} \).

Step 3: Apply the Limit Comparison Test

Consider the limit \( \lim_{n \to \infty} \frac{\frac{3}{n+\sqrt{n}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{3n}{n+\sqrt{n}} = \lim_{n \to \infty} \frac{3}{1+\frac{1}{\sqrt{n}}} = 3 \). This is a constant, implying that \( \sum_{n=1}^{\infty} \frac{3}{n+\sqrt{n}} \) has the same nature as \( \sum_{n=1}^{\infty} \frac{1}{n} \), which is known to diverge.

Key Concepts

Comparison Test

The Comparison Test is a handy tool when trying to determine the convergence or divergence of a series. This test involves comparing the series in question to another series whose convergence is already known. The idea is simple:

• To use the Comparison Test, one must find a series with positive terms that behaves like the series under question for large terms.
• If the known series converges and the series in question has smaller terms for all large indices, then it also converges.
• Conversely, if the known series diverges and the series in question has larger terms, then it diverges as well.

In the context of our exercise, we compared the series \( \ \ \sum_{n=1}^{\infty} \frac{3}{n+\sqrt{n}} \) to the harmonic series \( \ \ \sum_{n=1}^{\infty} \frac{1}{n} \), which is known to diverge. This comparison gives us insight about the behavior of the first series.

Limit Comparison Test

The Limit Comparison Test is a powerful technique that can offer more flexibility than the simple Comparison Test. It allows us to handle series where direct comparison is not as clear-cut. Here's how it works:

• We choose a second series with positive terms, similar to what we did with the Comparison Test.
• The next step is to compute the limit of the ratio of the terms of the series given and the chosen series.
• If this limit is a positive finite number, both series will either converge or diverge together.

In the example from the exercise, we computed \( \ \ \lim_{n \to \infty} \frac{\frac{3}{n+\sqrt{n}}}{\frac{1}{n}} \), which simplifies to 3. This shows that our series behaves similarly to the harmonic series, confirming its divergence.

Divergence of Series

Series divergence can sometimes be tricky to establish. A series is said to diverge when the sum of its infinite terms doesn't add up to a finite number. Recognizing divergence is crucial for understanding the broader behavior of a series. To determine divergence:

• Sometimes, analyzing the series directly can reveal divergence if the terms don't approach zero.
• Theorems like the Comparison Test and Limit Comparison Test are often needed for more complicated series.
• The Harmonic series, \( \ \ \sum_{n=1}^{\infty} \frac{1}{n} \), serves as a classic example of divergence, often used as a benchmark for other series.

In solving our exercise, realizing that our target series behaves like the harmonic series led us to conclude its divergence without too much computation. This strategic use of known divergent series is a basic yet effective approach to study convergence and divergence of more complex series.