Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{n=1}^{\infty} \frac{4^{n}+n}{n !}\)

Step-by-Step Solution

Answer

The series converges.

Step 1: Understand the Problem

The goal is to determine whether the infinite series \(\sum_{n=1}^{\infty} \frac{4^{n}+n}{n !}\) converges or diverges. The series has terms that combine an exponential function \(4^n\) and a linear term \(n\), divided by factorial \(n!\). We need to find a way to test for the convergence of this series.

Step 2: Split the Series

Notice that the series can be split into two separate series: \[\sum_{n=1}^{\infty} \frac{4^{n}+n}{n !} = \sum_{n=1}^{\infty} \frac{4^n}{n !} + \sum_{n=1}^{\infty} \frac{n}{n !}.\]This allows us to examine the convergence of each component separately.

Step 3: Examine the Exponential Term Series

The first series is \(\sum_{n=1}^{\infty} \frac{4^n}{n!}\). Recognize that this is a part of the exponential series expansion \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\). Specifically, for \(x = 4\), this series equals \(e^4 - \frac{4^0}{0!} = e^4 - 1\) because the series starts at \(n=1\). Therefore, it converges.

Step 4: Examine the Linear Term Series

The second series is \(\sum_{n=1}^{\infty} \frac{n}{n!}\). Consider the ratio test for this series. Calculate the limit:\[\lim_{n \to \infty} \left| \frac{(n+1)/(n+1)!}{n/n!} \right| = \lim_{n \to \infty} \left| \frac{n!n}{(n+1)n!} \right| = \lim_{n \to \infty} \frac{n}{(n+1)} = 1\].Since the limit is not less than 1, the ratio test is inconclusive. However, note that \(\frac{n}{n!}\) converges by the comparison test because it is much smaller than \(\frac{1}{n}\) for large \(n\).

Step 5: Confirm the Convergence of Each Part

Both parts \(\sum_{n=1}^{\infty} \frac{4^n}{n!}\) and \(\sum_{n=1}^{\infty} \frac{n}{n!}\) converge, as the exponential part follows the logic of \(e^x\) and the linear part converges by comparison to a simpler known series. Therefore, the whole series \(\sum_{n=1}^{\infty} \frac{4^n + n}{n!}\) converges.

Key Concepts

Exponential Series

The exponential series is a cornerstone of many mathematical concepts, serving as a fundamental expansion for exponential functions. It is represented by the formula:

• \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

This means that the exponential function can be expressed as an infinite sum of powers of the variable divided by factorials. This method of representation is advantageous, especially for calculations involving approximations of the exponential function.
In the context of the original exercise, one portion of the series is \(\sum_{n=1}^{\infty} \frac{4^n}{n!}\). This is a fragment of the exponential series, specifically for \(x = 4\). Therefore, by recognizing it as part of the series expansion of \(e^x\), we know that it converges. This convergence implies that the series approaches a specific value as \(n\) increases endlessly.

Factorial Denominator

Factorials are a mathematical concept that display rapid growth with increasing \(n\). A factorial \(n!\) is the product of all positive integers up to \(n\). For instance:

• \(1! = 1\)
• \(2! = 2\)
• \(3! = 6\)

This swift escalation means that factorial denominators contribute to the convergence of series because they make the terms vanish quickly as \(n\) grows. In the given exercise, both components of the series have \(n!\) in the denominator, strongly influencing the speed at which terms shrink. This is crucial in testing for convergence since smaller terms aid the series in converging.

Ratio Test

The ratio test is a popular method used to determine whether an infinite series converges or diverges. For a series \(\sum a_n\), the test involves always computing the limit:

• \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

The criteria for the test are:

• If the limit is less than 1, the series converges.
• If the limit is greater than 1, the series diverges.
• If the limit is exactly 1, the test is inconclusive.

In the given task, we applied the ratio test to \(\sum_{n=1}^{\infty} \frac{n}{n!}\). The limit evaluated to 1, which makes the test inconclusive. However, even though inconclusive, this test confirms that we need to explore other methods, such as the comparison test, to conclude convergence.

Comparison Test

The comparison test is a method to decide if a series converges by comparing it with another series whose convergence properties are known. In this test, we analyze a series \(\sum a_n\) by finding another series \(\sum b_n\) such that:

• If \(0 \leq a_n \leq b_n\) for all \(n\), and \(\sum b_n\) is convergent, then \(\sum a_n\) is also convergent.
• Conversely, if \(a_n \geq b_n\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

For the problem at hand, the series \(\sum_{n=1}^{\infty} \frac{n}{n!}\) was compared to \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), which is a well-known convergent series. Since \(\frac{n}{n!}\) falls below these terms for larger \(n\), it allows us to reason that the series is convergent by comparison to the convergent series \(\frac{1}{n^2}\). This comparison provided the needed evidence to establish convergence when the ratio test was inconclusive.