Problem 63 - A Graphical Approach to Precalculus with Limits

Question

Find each sum. $$\sum_{k=1}^{\infty}(0.3)^{k}$$

Step-by-Step Solution

Answer

The sum is \( \frac{3}{7} \).

Step 1: Recognize the Series Type

The given series \( \sum_{k=1}^{\infty}(0.3)^{k} \) is an infinite geometric series because it can be expressed as \( a + ar + ar^2 + ar^3 + \cdots \) where \( a = 0.3 \) and the common ratio \( r = 0.3 \).

Step 2: Check the Convergence Criterion

For an infinite geometric series to converge, the common ratio \( r \) must satisfy \( |r| < 1 \). In this case, \( |0.3| = 0.3 < 1 \), so the series converges.

Step 3: Use the Sum Formula for Infinite Geometric Series

The sum \( S \) of an infinite geometric series is given by the formula \( S = \frac{a}{1 - r} \). Here, \( a = 0.3 \) and \( r = 0.3 \).

Step 4: Substitute Values into the Formula

Substitute \( a = 0.3 \) and \( r = 0.3 \) into the sum formula: \[ S = \frac{0.3}{1 - 0.3} = \frac{0.3}{0.7}. \]

Step 5: Simplify the Expression

Calculate the simplified value of \( \frac{0.3}{0.7} \): Divide 0.3 by 0.7 to get \( S = \frac{3}{7} \).

Key Concepts

Infinite Series

An infinite series is a mathematical expression defined as the sum of an infinite sequence of terms. Imagine having a list of numbers that goes on forever and you are trying to add all of them. Infinite series might seem like they head toward an endless sum, but that is not always the case. In mathematics, it is possible for an infinite series to have a finite sum. This is one of the interesting aspects of infinite series, especially in the context of geometric series.

• Example of an infinite series: \(a + ar + ar^2 + ar^3 + \cdots \)

In this scenario, the terms are generated by multiplying a starting number \(a\) by a fixed number referred to as the "common ratio".

Convergence Criterion

The convergence criterion is essential for determining whether an infinite series approaches a finite value. Specifically for infinite geometric series, the convergence criterion hinges on the value of the common ratio \( r \).

• A series converges if the absolute value of \( r \) is less than 1: \(|r| < 1 \)
• If \( |r| \geq 1 \), the series does not converge.

For instance, in the series \( \sum_{k=1}^{\infty}(0.3)^{k} \), the common ratio \( r \) is \(0.3\), and since \( |0.3| < 1 \), the series meets the convergence criterion. In simple terms, the individual terms of the series become smaller and tend towards zero, allowing the sum to stabilize at a finite value.

Sum Formula

For a convergent infinite geometric series, there's a neat formula to find the sum. This formula is immensely helpful because it allows us to calculate the sum without having to add up endless terms manually. The sum \( S \) of an infinite geometric series is calculated using the formula:\[ S = \frac{a}{1 - r} \]where:

• \( a \) is the first term of the series
• \( r \) is the common ratio

To use this formula, just plug in the values for \( a \) and \( r \), so for our given series, we substitute \( a = 0.3 \) and \( r = 0.3 \) to get \( S = \frac{0.3}{1 - 0.3} = \frac{0.3}{0.7} = \frac{3}{7} \). That's the beauty of a formula - simplifying an infinite problem into an easy calculation!

Common Ratio

The common ratio in a geometric series is the factor by which we multiply each term to get the next one. It is a crucial element because it dictates the behavior of the series. If the common ratio is a fraction less than one, the series converges, leading to a finite sum. In the series \( \sum_{k=1}^{\infty}(0.3)^{k} \), the common ratio \( r \) is \(0.3\). Each term after the first is obtained by multiplying the previous term by \(0.3\).

• If \( r \) is positive and less than one: Series converges toward a finite sum.
• If \( r \) is greater than one: Series diverges, meaning it grows without bound.
• If \( r \) is negative: The terms oscillate in sign but can still converge if \(|r| < 1\).

Understanding the common ratio helps in quickly assessing whether an infinite series will converge and allows us to apply the sum formula effectively.