Problem 71
Question
In each of the geometric series in Exercises \(69-72,\) write out the first few terms of the series to find \(a\) and \(r,\) and find the sum of the series. Then express the inequality \(|r| < 1\) in terms of \(x\) and find the values of \(x\) for which the inequality holds and the series converges. $$ \sum_{n=0}^{\infty} 3\left(\frac{x-1}{2}\right)^{n} $$
Step-by-Step Solution
Verified Answer
The series converges when \(-1 < x < 3\), and the series sum is \(\frac{6}{3-x}\).
1Step 1: Identifying the First Terms
Firstly, identify the first few terms of the given series. The series is \( \sum_{n=0}^{\infty} 3\left(\frac{x-1}{2}\right)^{n} \). To find the terms, plug in values for \(n\): when \(n = 0\), the term is \(3\); when \(n = 1\), the term is \(3\left(\frac{x-1}{2}\right)\); when \(n = 2\), the term is \(3\left(\frac{x-1}{2}\right)^{2}\), and so on.
2Step 2: Identifying the Initial Term and Common Ratio
From the terms identified, we see that the initial term \(a\) is \(3\), and the common ratio \(r\) is \(\frac{x-1}{2}\).
3Step 3: Expressing Condition for Convergence
The series converges if the absolute value of the common ratio is less than one. So the inequality \(|r| < 1\) becomes \(\left| \frac{x-1}{2} \right| < 1\).
4Step 4: Solving the Inequality
Solve the inequality \(\left| \frac{x-1}{2} \right| < 1\). This leads to two inequalities: \( -1 < \frac{x-1}{2} < 1\). Multiply every part of the inequality by 2 to get \(-2 < x-1 < 2\). Finally, add 1 to all parts to obtain \(-1 < x < 3\).
5Step 5: Finding the Sum of the Series
Using the formula for the sum of an infinite geometric series, \( S = \frac{a}{1-r} \), where \(a = 3\) and \(r = \frac{x-1}{2}\), the sum is \( S = \frac{3}{1 - (\frac{x-1}{2})} = \frac{3}{1 - \frac{x-1}{2}} = \frac{3}{\frac{3-x}{2}} \). Simplifying, the sum is \(\frac{6}{3-x}\).
Key Concepts
Common RatioSeries ConvergenceSum of Series
Common Ratio
In a geometric series, each term is obtained by multiplying the previous term by a fixed, non-zero number called the _common ratio_.
For instance, if you have a series that begins with 3, the next term might be 3 times the common ratio, and the term after that would be the preceding term multiplied again by the common ratio. In the given problem, the geometric series is expressed as: \[ \sum_{n=0}^{\infty} 3\left(\frac{x-1}{2}\right)^{n} \] To identify the common ratio, examine the ratio of terms as you progress through the series. In this case, you will see that each term is multiplied by \( \frac{x-1}{2} \).
Thus, the common ratio, \( r \), is \( \frac{x-1}{2} \).
Understanding the common ratio is vital because it helps determine other properties of the series, such as whether the series converges or its total sum.
For instance, if you have a series that begins with 3, the next term might be 3 times the common ratio, and the term after that would be the preceding term multiplied again by the common ratio. In the given problem, the geometric series is expressed as: \[ \sum_{n=0}^{\infty} 3\left(\frac{x-1}{2}\right)^{n} \] To identify the common ratio, examine the ratio of terms as you progress through the series. In this case, you will see that each term is multiplied by \( \frac{x-1}{2} \).
Thus, the common ratio, \( r \), is \( \frac{x-1}{2} \).
Understanding the common ratio is vital because it helps determine other properties of the series, such as whether the series converges or its total sum.
Series Convergence
A geometric series converges if the terms approach a finite sum as they continue to infinity. This happens under a specific condition: the absolute value of the common ratio must be less than one.
In mathematical terms, for convergence, the condition is: \[ |r| < 1 \] Given the common ratio \( r = \frac{x-1}{2} \), ensure the series converges by solving the inequality: \[ \left| \frac{x-1}{2} \right| < 1 \] This inequality splits into two separate inequalities: - \( -1 < \frac{x-1}{2} < 1 \)Solving these provides the convergence interval for \( x \): - Multiply every part by 2: \( -2 < x-1 < 2 \)- Add 1 to all parts to get: \( -1 < x < 3 \) This means for \( x \) within the range of -1 and 3, the series converges to a finite sum.
In mathematical terms, for convergence, the condition is: \[ |r| < 1 \] Given the common ratio \( r = \frac{x-1}{2} \), ensure the series converges by solving the inequality: \[ \left| \frac{x-1}{2} \right| < 1 \] This inequality splits into two separate inequalities: - \( -1 < \frac{x-1}{2} < 1 \)Solving these provides the convergence interval for \( x \): - Multiply every part by 2: \( -2 < x-1 < 2 \)- Add 1 to all parts to get: \( -1 < x < 3 \) This means for \( x \) within the range of -1 and 3, the series converges to a finite sum.
Sum of Series
When a geometric series converges, we can calculate its sum even if it has infinitely many terms. The sum \( S \) of an infinite geometric series is given by the formula: \[ S = \frac{a}{1-r} \] where \( a \) is the first term and \( r \) is the common ratio.
In our exercise, the first term \( a = 3 \) and the common ratio \( r = \frac{x-1}{2} \). Plug these into the formula to find the sum: \[ S = \frac{3}{1 - \frac{x-1}{2}} \] To simplify this, rewrite the expression as: \[ S = \frac{3}{\frac{3-x}{2}} \] By multiplying both numerator and denominator by 2, you further simplify to: \[ S = \frac{6}{3-x} \] Thus, this simplified formula gives the sum of the series for \( x \) values within the convergence interval. Remember, using the sum formula is only valid when \( |r| < 1 \), ensuring the series is convergent.
In our exercise, the first term \( a = 3 \) and the common ratio \( r = \frac{x-1}{2} \). Plug these into the formula to find the sum: \[ S = \frac{3}{1 - \frac{x-1}{2}} \] To simplify this, rewrite the expression as: \[ S = \frac{3}{\frac{3-x}{2}} \] By multiplying both numerator and denominator by 2, you further simplify to: \[ S = \frac{6}{3-x} \] Thus, this simplified formula gives the sum of the series for \( x \) values within the convergence interval. Remember, using the sum formula is only valid when \( |r| < 1 \), ensuring the series is convergent.
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