Problem 36 - 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{n+2^{n}}{n^{2} 2^{n}}\end{equation}

Step-by-Step Solution

Answer

The series converges.

Step 1: Simplify the General Term

Start by simplifying the general term of the series \(a_n = \frac{n+2^n}{n^2 2^n}\). Separate the terms in the numerator to get \(a_n = \frac{n}{n^2 2^n} + \frac{2^n}{n^2 2^n}\). Simplifying gives \(a_n = \frac{1}{n \cdot 2^n} + \frac{1}{n^2}\).

Step 2: Analyze Each Part of the Series

Look at the two parts: \(b_n = \frac{1}{n \cdot 2^n}\) and \(c_n = \frac{1}{n^2}\).

Step 3: Determine Convergence of \(c_n\)

Recognize \(c_n = \frac{1}{n^2}\) as a p-series with \(p = 2 > 1\). A p-series with \(p > 1\) converges. Therefore, \(\sum \frac{1}{n^2}\) converges.

Step 4: Determine Convergence of \(b_n\)

For \(b_n = \frac{1}{n \cdot 2^n}\), compare with a known series. Use the comparison test, comparing it to \(d_n = \frac{1}{2^n}\), which is a geometric series with ratio \(\frac{1}{2}\). The geometric series \(\sum \frac{1}{2^n}\) converges. Since \(\frac{1}{n \cdot 2^n} \leq \frac{1}{2^n}\), by the comparison test, \(\sum b_n\) converges.

Step 5: Combine Results

Since both \(\sum b_n\) and \(\sum c_n\) converge individually, the overall series \(\sum_{n=1}^{\infty} \left( \frac{1}{n \cdot 2^n} + \frac{1}{n^2} \right)\) is the sum of two convergent series, which means the original series converges.

Key Concepts

p-series

A p-series is a type of infinite series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) where \( p \) is a positive constant. The convergence of a p-series depends on the value of \( p \). If \( p > 1 \), the p-series converges; however, if \( p \leq 1 \), it diverges.

For example, in the original exercise, the term \( c_n = \frac{1}{n^2} \) is a p-series with \( p = 2 \). Because \( 2 > 1 \), this series converges. Recognizing a p-series is crucial in determining the behavior of an infinite series and is a fundamental concept in calculus. Identifying p-series allows us to use known results about their convergence to aid in solving more complex series problems.

geometric series

A geometric series is a series where each term is a constant multiple of the previous term. It takes the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.

The convergence of a geometric series is straightforward:

• If \( |r| < 1 \), the series converges.
• If \( |r| \geq 1 \), the series diverges.

In the step-by-step solution, the term \( d_n = \frac{1}{2^n} \) was identified as part of a geometric series with a ratio \( r = \frac{1}{2} \). Since the ratio is less than 1, this series converges. Understanding geometric series is vital because they appear frequently in various mathematical contexts and provide a benchmark for analyzing other series through comparison.

comparison test

The comparison test is a method used to determine the convergence or divergence of an infinite series by comparing it to another series whose convergence is already known.

For this test, two variations are used:

• If \( a_n \leq b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( a_n \geq b_n \) for all \( n \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

In the exercise, \( b_n = \frac{1}{n \cdot 2^n} \) was compared to the geometric series \( \sum \frac{1}{2^n} \), a known convergent series. Since \( b_n \leq \frac{1}{2^n} \), the comparison test tells us that \( \sum b_n \) must also converge. This test is useful for evaluating relatively complex series by leveraging simpler known series.

infinite series

An infinite series is the sum of infinite terms and is expressed in the form \( \sum_{n=1}^{\infty} a_n \). The study of such series involves determining whether they add up to a finite value, which implies convergence, or do not, leading to divergence.

Key points about infinite series include:

• A series converges if the sequence of partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) approaches a finite limit as \( n \) approaches infinity.
• If the partial sums do not approach a finite limit, the series diverges.
• Various tests, such as the comparison test, p-series test, and geometric series analysis, are utilized to ascertain convergence or divergence.

In the context of the exercise, it is noted that the original series converges because it consists of the sum of two convergent series, \( \sum b_n \) and \( \sum c_n \). Understanding infinite series is fundamental to grasp various concepts in calculus and mathematical analysis because they provide insights into growth, summation, and limits at infinity.