Problem 34 - 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{\sqrt{n}}{n^{2}+1} \end{equation}

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^{2}+1}\) converges.

Step 1: Compare with Standard Series

To determine convergence or divergence, we need to compare the given series with a known p-series. The given series is \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^{2}+1} \). Notice that \( \sqrt{n} = n^{1/2} \). For large n, \( n^2+1 \approx n^2 \), so we approximate \( \frac{\sqrt{n}}{n^2+1} \approx \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}} \). This approximation resembles the p-series \( \sum \frac{1}{n^{p}} \) with \( p = \frac{3}{2} \).

Step 2: Determine Behavior of the Comparison Series

Recall that a p-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \), and diverges if \( p \leq 1 \). Here, \( p = \frac{3}{2} > 1 \), implying that the comparison p-series \( \sum \frac{1}{n^{3/2}} \) converges.

Step 3: Apply the Limit Comparison Test

To apply the Limit Comparison Test, consider the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n = \frac{\sqrt{n}}{n^2+1} \) and \( b_n = \frac{1}{n^{3/2}} \). Hence, \[ L = \lim_{n \to \infty} \frac{\frac{\sqrt{n}}{n^2+1}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{n^{3/2} \cdot \sqrt{n}}{n^2+1} = \lim_{n \to \infty} \frac{n^2}{n^2+1}. \]

Step 4: Evaluate the Limit

The limit \( L = \lim_{n \to \infty} \frac{n^2}{n^2+1} = \lim_{n \to \infty} \frac{1}{1+\frac{1}{n^2}} \) converges to 1 as \( n \to \infty \). Since \( 0 < L < \infty \), and since the comparison series \( \sum \frac{1}{n^{3/2}} \) converges, the Limit Comparison Test confirms that \( \sum \frac{\sqrt{n}}{n^2+1} \) also converges.

Key Concepts

p-series

A p-series is one of the simplest types of series we encounter in mathematics, particularly when studying convergence. A p-series is expressed as \( \sum \frac{1}{n^p} \), where \( p \) is a constant and \( n \) is a positive integer. Understanding the behavior of a p-series is key to analyzing many other series.

• If \( p > 1 \), the series converges. This means the sum approaches a finite number as we include more terms.
• If \( p \leq 1 \), the series diverges. In this case, the sum continues to grow without bound.

For example, in the solution, the series \( \sum \frac{1}{n^{3/2}} \) is identified as a p-series with \( p = \frac{3}{2} \). Since \( \frac{3}{2} > 1 \), it confirms that this p-series converges.

limit comparison test

The Limit Comparison Test is a powerful tool used to determine the convergence or divergence of a series by comparing it to another series whose behavior is known. Here's how it works:

• We have two series: \( \sum a_n \) and \( \sum b_n \), with terms \( a_n \) and \( b_n \) respectively.
• We calculate the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \).

Application and Evaluation

• If the limit \( 0 < L < \infty \), both series either converge or diverge together.
• If \( L = 0 \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( L = \infty \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

In the original exercise, the series \( \sum \frac{\sqrt{n}}{n^2+1} \) is analyzed using the Limit Comparison Test with \( \sum \frac{1}{n^{3/2}} \). By calculating \( L = 1 \), and knowing that the comparison series converges, we establish the convergence of the original series.

convergence and divergence of series

Determining when a series converges or diverges is a fundamental part of calculus and analysis. A convergent series is one where the sum of its terms approaches a specific finite value as more terms are added. Conversely, a divergent series is one where the sum does not approach any finite limit.

Key Indicators

• A quick way to judge convergence is by comparing a series to another well-known series, such as a p-series.
• Tests like the Limit Comparison Test provide a systematic method to evaluate series, especially when direct comparison is challenging.
• Understanding the underlying growth of terms, such as \( n^{1/2} \) approximately simplifying to a p-series, helps in identifying applicable tests.

In the problem, convergence of \( \sum \frac{\sqrt{n}}{n^2+1} \) was shown by comparing it to a convergent p-series. Each series offers unique insights and with the right tools, we can robustly determine their nature.