Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the sum \(S(x)\) of \(\sum_{n=0}^{\infty}(x-3)^{n}\). What is the convergence set?

Step-by-Step Solution

Answer

The sum is \( S(x) = \frac{1}{4-x} \), with a convergence set of \( 2 < x < 4 \).

Step 1: Understand the Expression

The expression given is an infinite geometric series: \( \sum_{n=0}^{\infty} (x-3)^{n} \). This type of series has a common ratio \( r \), which in this case is \( r = x-3 \).

Step 2: Identify the Formula for the Sum of Geometric Series

The sum \( S \) of an infinite geometric series \( \sum_{n=0}^{\infty} r^n \) can be found using the formula \( S = \frac{1}{1-r} \), given that \( |r| < 1 \).

Step 3: Apply the Formula

Substitute \( r = x-3 \) into the formula for the sum of the geometric series: \[ S(x) = \frac{1}{1-(x-3)} = \frac{1}{4-x} \]. This simplification comes from rearranging \( 1 - (x-3) \) to \( 4-x \).

Step 4: Determine the Convergence Set

The convergence set is determined by the condition \( |x-3| < 1 \), which ensures that the series converges. Solving \( |x-3| < 1 \) gives \( 2 < x < 4 \). Thus, the series converges for this interval.

Key Concepts

Convergence Interval

Understanding the convergence interval of a series is crucial. In an infinite geometric series like \( \sum_{n=0}^{\infty}(x-3)^{n} \), the convergence depends on the common ratio \( r \) being less than 1 in absolute value. Here, the common ratio is \( r = x-3 \). For this series to converge, the condition \( |x-3| < 1 \) must be met.

To solve for \( x \), we rearrange the inequality:

• First, express it as \( -1 < x-3 < 1 \).
• Add 3 to each part to isolate \( x \): \( 2 < x < 4 \).

So, the convergence interval for this series is \( 2 < x < 4 \). That means the values of \( x \) that make the series converge are those within this range. Understanding this interval helps in determining which specific values of \( x \) make calculations relevant to real-world problems.

Sum of Series

The sum of an infinite geometric series is a powerful concept in mathematics. To find the sum \( S(x) \) of the series \( \sum_{n=0}^{\infty}(x-3)^{n} \), we use the formula for the sum of an infinite geometric series: \( S = \frac{1}{1-r} \). This formula is applicable when \(|r| < 1\). Here, \( r = x-3 \).

After applying the formula, substituting \( x-3 \) for \( r \) gives us:

• \( S(x) = \frac{1}{1-(x-3)} \)
• Simplifying, this becomes \( S(x) = \frac{1}{4-x} \).

Calculating the sum in this way allows for an elegant solution that gives a finite value for \( S(x) \) over the convergence interval. It's especially useful in summing series that might originate in various applications, such as physics or finance, where such converge is necessary.

Geometric Series Formula

The geometric series formula is a fundamental tool in mathematics for summing series. For an infinite geometric series like \( \sum_{n=0}^{\infty} r^n \), the general sum formula is \( S = \frac{1}{1-r} \) given that \(|r| < 1\). This formula provides a simple way to find the sum when the series converges.

Understanding how to apply this formula is beneficial:

• Identify the common ratio, \( r \).
• Ensure \(|r| < 1\) for convergence.
• Use \( S = \frac{1}{1-r} \) to find the sum.

In the case of our example with \( r = x-3 \), inserting this into the formula correctly yields the sum. The geometric series formula is a shortcut that uses the power of algebra to provide answers without needing to calculate each term separately, showcasing the beauty and efficiency of mathematical series applications.