Problem 16 - Essential Calculus Early Transcendentals

Question

Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \frac{1+3^{n}}{2^{n}}$$

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Break Down the Series

The given series is \(\sum_{n=1}^{\infty} \frac{1+3^{n}}{2^{n}}\). We can break this series into two separate series: \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) and \(\sum_{n=1}^{\infty} \frac{3^n}{2^n}\). This helps simplify the process of determining convergence and finding sums.

Step 2: Evaluate Convergence of the First Series

Consider the first series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\). This is a geometric series with a common ratio \(r = \frac{1}{2}\), which is less than 1. Therefore, it is convergent.

Step 3: Find the Sum of the First Series

For the convergent geometric series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\), the sum is given by \(\frac{a}{1-r}\), where \(a = \frac{1}{2}\) is the first term and \(r = \frac{1}{2}\). So the sum is \(\frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1\).

Step 4: Evaluate Convergence of the Second Series

Consider the second series \(\sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^n\). This is also a geometric series, but with a common ratio \(r = \frac{3}{2}\), which is greater than 1. Therefore, this series diverges.

Step 5: Determine Overall Convergence

The original series \(\sum_{n=1}^{\infty} \frac{1+3^n}{2^n}\) is the sum of a convergent series and a divergent series. Since the divergent series dominates, the original series is divergent.

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. It's a fascinating concept because it allows us to find the sum of infinite terms under certain conditions.

An infinite geometric series is expressed as \[S = a + ar + ar^2 + ar^3 + \cdots \]where:

• \(a\) is the first term of the series
• \(r\) is the common ratio

To determine if a geometric series converges, it's important to look at the common ratio \(r\). If the absolute value of \(r\) is less than 1 (\(|r| < 1\)), the series converges and we can find its sum using the formula \(S = \frac{a}{1-r}\).

If \(|r| \geq 1\), the series does not converge, meaning the partial sums continue to grow indefinitely.

Convergent Series

Understanding when a series converges is crucial in mathematics. A series converges if the sum of its infinite terms approaches a finite number.

For example, in a geometrical series with a common ratio \(r\) meeting the condition \(|r| < 1\), the sum of the infinite series can be calculated using the formula \[S = \frac{a}{1-r}.\]

This concept is illustrated with the series given by \(\sum_{n=1}^{\infty} \frac{1}{2^n}\). Here, we have each term getting successively smaller and closer to zero, with \(\frac{1}{2}\) as its ratio. This makes the series not only approachable but gives it a precise sum, 1 in this case.

An important aspect of convergent series is that they have a finite sum, making them bounded and predictable. This property is useful in mathematical calculations and problem solving, especially when infinite series appear.

Divergent Series

In contrast to convergent series, a divergent series is one where the sum of its terms does not settle to any finite limit. Instead, these sums grow without bound as more terms are added.

For example, in the series \(\sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^n\), the common ratio is \(\frac{3}{2}\), which is greater than 1. With such a ratio, each term becomes progressively larger, leading the partial sums to increase indefinitely.

Thus, this series diverges. Divergent series can be useful for analyzing limits and behaviors within sequences, even if they lack a definite sum. Though they do not converge to a single number, divergent series still play a significant role in calculus and when exploring the theoretical limits of functions and operations.