Problem 45 - College Algebra with Modeling and Visualization

Question

Find the sum of the infinite geometric series. $$ 1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\cdots $$

Step-by-Step Solution

Answer

The sum of the series is \(\frac{3}{2}\).

Step 1: Identify the Series Type

The series given is: \[1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\]This is an infinite geometric series since every term after the first is obtained by multiplying the previous term by a constant ratio.

Step 2: Determine the First Term and Common Ratio

In a geometric series, the first term is denoted by \(a\) and the common ratio by \(r\). Here, the first term \(a = 1\) and the common ratio \(r\) is calculated as the ratio between any two successive terms. For example, \[ r = \frac{\frac{1}{3}}{1} = \frac{1}{3} \] Thus, \(r = \frac{1}{3}\).

Step 3: Check the Convergence Criterion

The sum of an infinite geometric series converges if the absolute value of the common ratio is less than 1. Check if this is true:\[ |r| = \left|\frac{1}{3}\right| = \frac{1}{3} < 1 \] Hence, the series converges.

Step 4: Apply the Infinite Geometric Series Sum Formula

The sum \(S\) of an infinite geometric series is calculated using the formula:\[ S = \frac{a}{1 - r} \]Substituting the known values \(a = 1\) and \(r = \frac{1}{3}\), the sum becomes:\[ S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \]

Step 5: Verify the Calculations

Recalculate the sum using the same formula to ensure there are no mistakes:\[ S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \]Both calculations give the sum \(\frac{3}{2}\), confirming the result is correct.

Key Concepts

Sum of Series

An infinite geometric series allows us to find the sum of infinitely many terms, provided the series converges. This might seem like a challenging idea since you're technically adding up an endless number of terms. However, there's a handy formula to find this sum, making it easier than you might think.
The essential formula used here is: \[ S = \frac{a}{1 - r} \] Where:

• \( S \) is the sum of the series,
• \( a \) is the first term of the series,
• \( r \) is the common ratio.

This formula works because each subsequent term gets smaller and smaller when the common ratio is between -1 and 1, thus overall representing a fraction of the first term.

Geometric Sequence

A geometric sequence is characterized by each term being a constant multiple of the preceding term. In simpler terms, you get each term by multiplying the one before it by the same ratio. This ratio, called the common ratio \( r \), defines the sequence's nature.
Identifying the first term and common ratio is crucial:

• The first term, \( a \), is the starting point of the sequence.
• The common ratio, \( r \), is found by dividing one term by its previous term, for example, \( r = \frac{1}{3} \) in our sequence.

This consistent multiplication by \( r \) forms a sequence that can either grow infinitely or shrink towards zero, depending on the value of \( r \). A geometric sequence can be finite or infinite, with the latter being the focus in this case.

Series Convergence

Convergence in the context of a series means that adding all the terms up results in a finite number. This is an essential characteristic for understanding if an infinite geometric series can be summed.
A series converges when the absolute value of its common ratio \( r \) is less than 1. This is because, with each multiplication by \( r \) (as long as \(|r|<1\)), the terms will continually decrease in size. As they get smaller and smaller, their collective sum approaches a specific value, rather than increasing indefinitely.
In the given exercise, \(|r| = \frac{1}{3}\), which is indeed less than 1. This guarantees that our infinite geometric series converges, allowing us to properly calculate its sum. Simply put, if the terms keep getting smaller due to \(|r|<1\), we can neatly find a sum that is finite.