Problem 13 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Do the following. i. Write out the rst two terms of the series. ii. Determine whether or not the series converges. iii. If the series converges, determine its sum. $$ \sum_{n=2}^{\infty} \frac{3^{n}}{4^{n-1}} $$

Step-by-Step Solution

Answer

The first two terms of the series are \(\frac{9}{4}\) and \(\frac{27}{16}\). The series converges because the common ratio is \(\frac{3}{4}\), which is less than 1 in absolute value. The sum of the series is 9.

Step 1: Writing Out The First Two Terms

The first term is when \(n=2\), so it's \(\frac{3^{2}}{4^{2-1}} = \frac{9}{4}\). The second term is when \(n=3\), so it’s \(\frac{3^{3}}{4^{3-1}} = \frac{27}{16}\).

Step 2: Determining Whether or Not The Series Converges

The common ratio of the series is \(\frac{3}{4}\). The series converges because the absolute value of the common ratio is less than 1.

Step 3: Determining The Sum Of The Series

Since the series is convergent, the sum is given by \(S_{\infty} = \frac{a}{1 - r}\), where \(a\) is the first term (\(\frac{9}{4}\)) and \(r\) is the common ratio (\(\frac{3}{4}\)). So the sum is \(S_{\infty} = \frac{\frac{9}{4}}{1 - \frac{3}{4}} = \frac{9}{1}=9.\)

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. In mathematical terms, a geometric series can be expressed as \[ \sum_{n=1}^{\infty} ar^{n-1} \] where \( a \) is the first term and \( r \) is the common ratio.

• Each term is generated by multiplying the preceding term by \( r \).
• This kind of series can either be finite or infinite.

For example, in the series \( \sum_{n=2}^{\infty} \frac{3^{n}}{4^{n-1}} \), we see that after taking out the first two terms, each subsequent term is produced by multiplying the previous term by \( \frac{3}{4} \). This represents a geometric progression.

Common Ratio

The common ratio in a geometric series is the factor by which we multiply each term to obtain the next term. It is a crucial part in determining the behavior of the series.

• Calculated by dividing any term by its preceding term.
• If \(|r| < 1\), the terms become successively smaller, directing towards zero.
• If \(|r| > 1\), the terms grow larger, potentially without bound.

In our series example, the common ratio \( r \) is \( \frac{3}{4} \). Since \(|\frac{3}{4}| < 1\), this series shrinks in size and approaches zero as the series progresses, indicating that it will converge.

Infinite Series

An infinite series is the sum of the terms of an infinite sequence. Unlike a finite series which has a limited number of terms, an infinite series does not have an end.

• It involves adding up numbers forever, but practically, we look at limits and convergence.
• The series can diverge or converge.
• Convergence depends largely on the common ratio for geometric series.

In our example, \( \sum_{n=2}^{\infty} \frac{3^{n}}{4^{n-1}} \) is an infinite series because the upper limit of \( n \) is infinity, implying endless terms. We are interested in knowing if the series sums to a finite number or not, by using convergence properties.

Convergent Series

A series converges when the sum of its infinite terms approaches a specific limit as you continue to add more terms. When dealing with geometric series, we typically look at the common ratio to establish convergence.

• If the absolute value of the common ratio \(|r| < 1\), the series converges.
• If \(|r| \geq 1\), the series diverges.
• The sum of a convergent geometric series \( S_\infty \) can be calculated using the formula: \[ S_\infty = \frac{a}{1 - r} \]

For the series \( \sum_{n=2}^{\infty} \frac{3^{n}}{4^{n-1}} \), since \( r = \frac{3}{4} \) and \(|\frac{3}{4}| < 1\), it is convergent. The sum of this series is found using the formula, giving us a finite sum of 9.