Problem 36 - Algebra 2

Question

Evaluate each infinite series that has a sum. $$ \sum_{n=1}^{\infty}\left(-\frac{1}{3}\right)^{n-1} $$

Step-by-Step Solution

Answer

The sum of the series \(\sum_{n=1}^{\infty}\left(-\frac{1}{3}\right)^{n-1}\) is \(\frac{3}{4}\).

Step 1 - Identify the first term and the ratio

The first term 'a' in the series is 1 (since the series starts with \(n=1\) and the base of the exponent is \(-\frac{1}{3}\)). The ratio 'r' is \(-\frac{1}{3}\). So we have that \(a = 1\) and \(r = -\frac{1}{3}\).

Step 2 - Check whether the series converges

For the geometric series to converge, the absolute value of the ratio |r| must be less than 1. Here, \(|-\frac{1}{3}| = \frac{1}{3}\), which is less than 1. So, the series converges and has a sum.

Step 3 - Use the formula to calculate the sum

The formula to find the sum of an infinite geometric series is \(S = \frac{a}{1-r}\). Substituting the values for a and r from step 1 into the formula, the sum \(S = \frac{1}{1-\left(-\frac{1}{3}\right)} = \frac{1}{1+\frac{1}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4}\).

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. This type of series is foundational in understanding complex mathematical concepts, yet simple to grasp once you break it down. In our given problem, the series is \[ \sum_{n=1}^{\infty}\left(-\frac{1}{3}\right)^{n-1} \] Here, the term when \( n=1 \) gives us our first term, which is 1.

• First Term (a): This is the starting point of the series. In our example, it's 1.
• Common Ratio (r): This is the factor by which we multiply each term to get the next one. Here, it's given as \(-\frac{1}{3}\).

These elements help form the building blocks of the geometric series. Understanding these components allows us to analyze the behavior of the series, especially when studying if it will converge.

Convergence of Series

In the context of infinite series, convergence refers to whether the series adds up to a finite value. This is particularly interesting when dealing with geometric series, as they may converge or diverge based on the value of the common ratio \( |r| \).
For the series to converge, \( |r| \) must be less than 1. This tells us that the terms in the series are getting smaller and smaller, eventually approaching zero. In our problem, the series \[ \sum_{n=1}^{\infty}\left(-\frac{1}{3}\right)^{n-1} \] has a ratio \( r = -\frac{1}{3} \). Since \( |-\frac{1}{3}| = \frac{1}{3} \), which is less than 1, the series converges.

• Absolute Value Condition: The series will converge if \( |r| < 1 \).
• Approaching Zero: As the terms get smaller, they contribute less and less to the sum, allowing the series to settle at a finite value.

Convergence is crucial because it indicates whether an infinite series has a practical sum.

Sum of Series

The sum of an infinite geometric series, when it converges, can be calculated using a specific formula. This formula is vital as it allows us to find the total sum without manually adding an infinite number of terms. The formula for finding the sum \( S \) of a convergent infinite geometric series is \[ S = \frac{a}{1-r} \] where \( a \) is the first term and \( r \) is the common ratio.
In our exercise, the first term \( a = 1 \) and the common ratio \( r = -\frac{1}{3} \). Plugging these values into the formula gives us \[ S = \frac{1}{1 - (-\frac{1}{3})} = \frac{1}{1 + \frac{1}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \].

• Formula Application: Utilize the sum formula for efficient calculation.
• Simplicity in Calculation: This approaches provides a clear, formulaic way to solve for the sum.

Using this approach ensures that even infinite series become manageable and comprehensible.