Problem 59 - Calculus: Early Transcendentals

Question

Find the values of \( x \) for which the series converges. Find the sum of the series for those values of \( x. \) \( \displaystyle \sum_{n = 0}^{\infty} \frac {(x - 2)^n}{3^n} \)

Step-by-Step Solution

Answer

Converges for \(-1 < x < 5\), sum is \(\frac{3}{5-x}\).

Step 1: Recognize the Series

The given series is \( \sum_{n=0}^{\infty} \frac{(x-2)^n}{3^n} \). This series is in the form of a geometric series \( \sum_{n=0}^{\infty} ar^n \) where \( a = 1 \) and the common ratio \( r = \frac{x-2}{3} \).

Step 2: Determine Convergence Condition

A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if the absolute value of the common ratio \( |r| < 1 \). Here, \( | \frac{x-2}{3} | < 1 \).

Step 3: Solve the Inequality for Convergence

Solve \( | \frac{x-2}{3} | < 1 \) to find the interval of convergence. The inequality \( | \frac{x-2}{3} | < 1 \) means \( -1 < \frac{x-2}{3} < 1 \).

Step 4: Simplify the Inequality

Multiply all parts of the inequality \( -1 < \frac{x-2}{3} < 1 \) by 3 to isolate \( x-2 \): \(-3 < x - 2 < 3\).

Step 5: Solve for x

Add 2 to each part of the inequality \(-3 < x - 2 < 3\) to solve for \( x \): \(-1 < x < 5\). This is the interval of convergence for \( x \).

Step 6: Calculate the Sum of the Series

For a converging geometric series \( \sum_{n=0}^{\infty} ar^n \), the sum is \( \frac{a}{1-r} \). Here, the sum is \( \frac{1}{1 - \frac{x-2}{3}} \).

Step 7: Simplify the Sum Expression

Simplify \( \frac{1}{1 - \frac{x-2}{3}} \) to find the sum of the series in terms of \( x \). It becomes \( \frac{3}{5-x} \).

Step 8: State the Final Results

The series converges for \( -1 < x < 5 \), and the sum of the series for these values of \( x \) is \( \frac{3}{5-x} \).

Key Concepts

Series Convergence

In mathematics, a series is an expression that shows the sum of the terms of a sequence. The convergence of a series is a central concept because it lets us know whether a series will lead to a finite sum or not.
For geometric series—such as the one given in the problem \( \sum_{n=0}^{\infty} \frac{(x-2)^n}{3^n} \)—the rule for convergence is quite straightforward. A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if the absolute value of its common ratio \( |r| \) is less than 1.
This condition \( |r| < 1 \) ensures that the terms of the series get progressively smaller, approaching zero, such that the infinite sum stabilizes at a certain number rather than diverging to infinity.
In this case, the common ratio \( r = \frac{x-2}{3} \), so the convergence condition becomes \{|\frac{x-2}{3}| < 1\}. When this inequality is satisfied, the series converges.

Interval of Convergence

The interval of convergence is the range of \( x \) values for which the series converges. For the given series, we have the inequality \( |\frac{x-2}{3}| < 1 \), which can be solved to determine this range.

• Start by multiplying each part of the inequality \( -1 < \frac{x-2}{3} < 1 \) by 3 to simplify: \(-3 < x-2 < 3\).
• Then, solve for \( x \) by adding 2 throughout this inequality: \(-1 < x < 5\). This results in the interval \(-1 < x < 5\).

This interval represents all the \( x \) values for which the series will converge to a finite value. Outside this range, the series diverges, meaning it does not approach a specific number.

Sum of Series

Once it's established that a series converges, you can find its sum. For a converging geometric series \( \sum_{n=0}^{\infty} ar^n \), the sum is calculated using the formula \( \frac{a}{1-r} \).
In the series given in the problem, the first term \( a = 1 \) and the common ratio \( r = \frac{x-2}{3} \).

• Plugging these into the sum formula gives: \( \frac{1}{1 - \frac{x-2}{3}} \).
• To simplify this expression, multiply the numerator and the denominator by 3: \( \frac{3}{3 - (x-2)} = \frac{3}{5-x} \).

This formula, \( \frac{3}{5-x} \), is the sum of the series when \( x \) is within the interval of convergence, \(-1 < x < 5\). It succinctly summarizes the series' overall behavior within its convergent range.