Problem 75 - Calculus: Early Transcendentals

Question

Find the value of \( c \) if \( \displaystyle \sum_{n = 2}^{\infty} (1 + c)^{-n} = 2 \)

Step-by-Step Solution

Answer

The value of \( c \) is \( \frac{-1 + \sqrt{3}}{2} \).

Step 1: Understand the Series

The problem involves an infinite geometric series. An infinite geometric series is given by the formula \( \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \) for \(|r|<1\). Here, \( a = (1+c)^{-2} \) and \( r = \frac{1}{1+c} \).

Step 2: Apply the Infinite Geometric Series Formula

The sum of an infinite geometric series is \( \frac{a}{1-r} \). For this series, \( a = (1+c)^{-2} \) and \( r = \frac{1}{1+c} \), giving us the equation: \[ \frac{(1+c)^{-2}}{1 - \frac{1}{1+c}} = 2. \]

Step 3: Simplify the Denominator

Simplify the denominator: \( 1 - \frac{1}{1+c} = \frac{1+c-1}{1+c} = \frac{c}{1+c} \). The equation becomes: \[ (1+c)^{-2} \times \frac{1+c}{c} = 2. \]

Step 4: Simplify and Solve for \( c \)

Continue the simplification: \( \frac{1}{c(1+c)} = 2 \), which leads to \( c(1+c) = \frac{1}{2} \). Thus, \( c^2 + c - \frac{1}{2} = 0 \).

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = 1 \), and \( c = -\frac{1}{2} \). Compute: \[ c = \frac{-1 \pm \sqrt{1 + 2}}{2} = \frac{-1 \pm \sqrt{3}}{2}. \]

Step 6: Validate the Solution Range

Since the series converges if \( |r|<1 \), \( \frac{1}{1+c} < 1 \) implies \( c > 0 \). Thus, the valid value for \( c \) is \( \frac{-1 + \sqrt{3}}{2} \).

Key Concepts

Quadratic Equation

A quadratic equation is a vital part of algebra. It is any equation that can be written in the standard form: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In this exercise, we derived a quadratic equation from the simplification and manipulation of the series expression.

• Quadratic equations can have two solutions, one solution, or no real solutions at all.
• The solutions of a quadratic equation can be computed using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our example, the quadratic equation \( c^2 + c - \frac{1}{2} = 0 \) was solved by substituting \( a = 1 \), \( b = 1 \), and \( c = -\frac{1}{2} \) into the quadratic formula. This yields two potential solutions, but further conditions help determine which solution fits the problem's requirements.

Series Convergence

Understanding series convergence is crucial when dealing with infinite series. An infinite series converges if the sum of its terms approaches a specific value as the number of terms increases towards infinity.

• The series in question here is an infinite geometric series, which converges if the absolute value of the common ratio \( |r| \) is less than 1.
• This is fundamentally important because, without convergence, the series would not have a finite sum.

In this exercise, the condition \( |r| < 1 \) leads to the realization that \( \frac{1}{1+c} < 1 \), which subsequently provides valuable constraints that determine the feasibility of the solution for \( c \). If the series converges, it means the value of \( c \) we derive keeps them in this valid region.

Geometric Series Formula

A geometric series is one where each term constitutes a constant multiple, called the common ratio \( r \), of the previous term. The formula used to sum an infinite geometric series is key.

• The formula \( S = \frac{a}{1-r} \) applies if the series converges, meaning \( |r| < 1 \).
• Here, \( a \) represents the first term of the series, and \( r \) represents the common ratio.

In this specific scenario, we applied the geometric series formula by identifying \( a = (1+c)^{-2} \) and \( r = \frac{1}{1+c} \). By substituting these values into the formula, we formulated the equation \( \frac{(1+c)^{-2}}{1 - \frac{1}{1+c}} = 2 \), which was instrumental for solving for \( c \). This formula thus assists in linking the sum of an infinite series directly with its terms and ratio.

Sum of an Infinite Series

The sum of an infinite series is the limit that the partial sums of the series approach as the number of terms tends to infinity. Calculating this sum involves understanding convergence and the specific series formula.

• For an infinite geometric series, when it converges, the sum is given by \( S = \frac{a}{1-r} \).
• This formula is applicable provided \( |r| < 1 \), ensuring that the terms decrease rapidly enough to zero, hence forming a finite sum.

In the exercise, we have the series starting from \( n=2 \) onwards, denoted by \( \sum_{n=2}^{\infty} (1+c)^{-n} = 2 \). By applying the geometric series sum formula correctly, we were effectively able to find the value of \( c \) that satisfies this equation. This procedure shows how key skills in manipulating and understanding formulas can lead to finding meaningful results in mathematical problems.