Problem 5 - Calculus: One and Several Variables

Question

Find the sum of the series. $$\sum_{k=0}^{\infty} \frac{3}{10^{k}}$$

Step-by-Step Solution

Answer

The sum of the given infinite geometric series is \(\frac{10}{3}\).

Step 1: Identify the first term, a

The first term of the series is obtained by plugging k=0 into the general term formula \(\frac{3}{10^k}\). \(a = \frac{3}{10^0} = \frac{3}{1} = 3\).

Step 2: Identify the common ratio, r

The common ratio between consecutive terms is the factor by which each term can be multiplied to obtain the following term. In this series, divide term k+1 by term k: $$r = \frac{3}{10^{k+1}} \cdot \frac{10^k}{3} = \frac{3 \cdot 10^k}{3 \cdot 10^{k+1}} = \frac{1}{10}$$ So, the common ratio r = \(\frac{1}{10}\).

Step 3: Use the sum formula for an infinite geometric series

Now, apply the sum formula for an infinite geometric series: $$S = \frac{a}{1 - r}$$ Substitute the values a = 3 and r = \(\frac{1}{10}\) into the formula: $$S = \frac{3}{1 - \frac{1}{10}} = \frac{3}{\frac{9}{10}} = \frac{3 \cdot 10}{9} = \frac{30}{9}$$

Step 4: Simplify the fraction

Simplify the fraction and obtain the result: $$S = \frac{30}{9} = \frac{10}{3}$$ Hence, the sum of the given infinite geometric series is \(\frac{10}{3}\).

Key Concepts

Sum of Series

An **infinite geometric series** is a sequence of numbers where each term is found by multiplying the previous term by a fixed, non-zero number called the common ratio. The sum of an infinite geometric series can be calculated if the absolute value of the common ratio is less than 1. In this exercise, we are given the series:\[ \sum_{k=0}^{\infty} \frac{3}{10^{k}} \]The goal is to find the sum of this series.To achieve that, we need to use the formula for the sum, \( S \), of an infinite geometric series:\[ S = \frac{a}{1 - r} \]where \( a \) is the first term and \( r \) is the common ratio of the series. By substituting the first term and the common ratio into this formula, we are able to find the sum. In this particular example:

• First term, \( a = 3 \)
• Common ratio, \( r = \frac{1}{10} \)

Substituting into the sum formula, we get:\[ S = \frac{3}{1 - \frac{1}{10}} = \frac{3}{\frac{9}{10}} = \frac{30}{9} = \frac{10}{3} \]Thus, the sum of this infinite geometric series is \( \frac{10}{3} \).

Common Ratio

The **common ratio** is the factor by which consecutive terms of a geometric sequence multiply to obtain the next term. Identifying the common ratio is crucial for determining the behavior of the series.In our example series:\[ \sum_{k=0}^{\infty} \frac{3}{10^{k}} \]The common ratio \( r \) is calculated by dividing any term in the sequence by the previous term. For our geometric progression, each term can be written as \( \frac{3}{10^k} \). So, the common ratio is determined by:\[ r = \frac{\frac{3}{10^{k+1}}}{\frac{3}{10^{k}}} = \frac{1}{10} \]This shows that each term is \( \frac{1}{10} \) of the preceding term. If \(|r| < 1\), which in this case is true, the series will converge towards a sum. This enables us to use the sum formula for an infinite geometric series.

Geometric Progression

A **geometric progression** (or sequence) is a series of numbers each found by multiplying the previous one by a fixed, non-zero number: the common ratio. This makes identification of the common ratio paramount, as it can give insight into the nature and eventual sum of the progression, if applicable.In the series stated in this exercise:\[ \sum_{k=0}^{\infty} \frac{3}{10^{k}} \]we identify it as a geometric progression because:

• The first term is \( a = 3 \).
• Each subsequent term is multiplied by the common ratio, \( r = \frac{1}{10} \).

Each number can be expressed explicitly in the form \( \frac{3}{10^k} \), where \( k \) is an integer starting from 0. This repetitive pattern defines it as a geometric progression. Understanding this structure is key for solving such sequences efficiently. Remember, if the common ratio is less than 1, the series will steadily approach a finite sum, as demonstrated in this problem.