Question

Consider the following infinite series. a. Find the first four terms of the sequence of partial sums. b. Use the results of part (a) to find a formula for \(S_{n}\) c. Find the value of the series. $$\sum_{k=1}^{\infty} \frac{2}{3^{k}}$$

Short answer

Answer: The value of the series is 1.

Step-by-step solution

Identify the General Term#a_n\( of the Series

We have an infinite series given by $$\sum_{k=1}^{\infty} \frac{2}{3^{k}}$$ The general term of the series can be written as \)a_k = \frac{2}{3^k}$.

Find the First Four Terms of the Sequence

Calculate the first four terms of the sequence by substituting the values of \(k\) (1, 2, 3, and 4) into the general term \(a_k\). \(k=1: a_1 = \frac{2}{3^1} = \frac{2}{3}\) \(k=2: a_2 = \frac{2}{3^2} = \frac{2}{9}\) \(k=3: a_3 = \frac{2}{3^3} = \frac{2}{27}\) \(k=4: a_4 = \frac{2}{3^4} = \frac{2}{81}\) The first four terms of the sequence are \(\frac{2}{3}, \frac{2}{9}, \frac{2}{27}, \frac{2}{81}\).

Find the First Four Partial Sums

Calculate the first four partial sums by adding up the first n terms (\(S_n\)). \(S_1 = a_1 = \frac{2}{3}\) \(S_2 = a_1 + a_2 = \frac{2}{3} + \frac{2}{9} = \frac{8}{9}\) \(S_3 = a_1 + a_2 + a_3 = \frac{2}{3} + \frac{2}{9} + \frac{2}{27} = \frac{26}{27}\) \(S_4 = a_1 + a_2 + a_3 + a_4 = \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} = \frac{80}{81}\)

Find a Formula for \(S_n\)

To find a formula for \(S_n\), notice that the series is a geometric series, with a common ratio of \(\frac{1}{3}\). We can use the geometric series formula for this: $$S_n = \frac{a_1(1-r^n)}{1-r}$$ where \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. In this case, \(a_1 = \frac{2}{3}\) and \(r = \frac{1}{3}\). Substitute this into the equation to get the formula for \(S_n\): $$S_n = \frac{\frac{2}{3}(1-\left(\frac{1}{3}\right)^n)}{1-\frac{1}{3}}$$

Find the Value of the Series

To find the value of the series, we need to find the limit of \(S_n\) as n approaches infinity: $$\lim_{n\to\infty} S_n = \lim_{n\to\infty} \frac{\frac{2}{3}(1-\left(\frac{1}{3}\right)^n)}{1-\frac{1}{3}}$$ Since the common ratio \(r\) is between -1 and 1 (-1<\(\frac{1}{3}\)<1), the limit converges to: $$\lim_{n\to\infty} (1-\left(\frac{1}{3}\right)^n) = 1$$ Hence, the value of the series is: $$\sum_{k=1}^{\infty} \frac{2}{3^{k}} = \frac{\frac{2}{3}(1)}{1-\frac{1}{3}} = \boxed{1}$$

Key concepts

Partial Sums

When we discuss partial sums in the context of an infinite series, we are essentially considering a sequence of sums. Each sum, called a partial sum, is a finite sum of the first few terms of the series.
Let's break this down:

• First Term: The first partial sum, denoted as \( S_1 \), is simply equal to the first term of the series. For example, given an infinite series like \( \sum_{k=1}^{\infty} \frac{2}{3^{k}} \), the first term \( a_1 \) is \( \frac{2}{3} \), so our first partial sum \( S_1 = \frac{2}{3} \).
• Subsequent Terms: To find \( S_2 \), you add the first two terms together: \( S_2 = a_1 + a_2 = \frac{2}{3} + \frac{2}{9} = \frac{8}{9} \). This process repeats for more terms: \( S_3 = \frac{2}{3} + \frac{2}{9} + \frac{2}{27} = \frac{26}{27} \) and \( S_4 = \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} = \frac{80}{81} \).

Each partial sum gives us a closer approximation to the total sum of the series. This concept of partial sums becomes pivotal in understanding whether a series converges or diverges.

Convergence of Series

In mathematics, particularly in series, convergence refers to whether a series approaches a specific value as you add infinitely many terms. We can test convergence by observing the sequence of partial sums.
Convergence happens when the sequence of partial sums gets closer and closer to a particular value, even as the number of terms (or \( n \)) goes to infinity. Consider the infinite geometric series \( \sum_{k=1}^{\infty} \frac{2}{3^{k}} \):

• Behavior of Terms: As \( n \) increases, each term \( a_n = \frac{2}{3^n} \) becomes smaller, meaning the new terms contribute less to the partial sums.
• Limit of Partial Sums: We find the limit of the partial sums \( \lim_{n\to\infty} S_n \). For our series, this limit equals 1, indicating convergence to this specific value.

The critical factor in determining convergence is the common ratio. If the absolute value of this ratio \( |r| \) is less than 1, the series will converge. In our example, the common ratio is \( \frac{1}{3} \), which satisfies this condition, thus the series converges to 1.

Geometric Series Formula

The geometric series formula is a powerful tool for analyzing and evaluating the sums of both finite and infinite geometric series. It provides a simple way to express the sum of a series based on its first term, common ratio, and number of terms.
For a geometric series with first term \( a_1 = \frac{2}{3} \) and common ratio \( r = \frac{1}{3} \), the sum of the first \( n \) terms, \( S_n \), is calculated using:

• The formula is given by \( S_n = \frac{a_1(1-r^n)}{1-r} \).
• The numerator represents the product of the first term and a term that decreases as \( n \) increases, given by \( 1-r^n \).
• The denominator, \( 1-r \), ensures we are dividing by the complement of the ratio, leading to simplified results.

When \( n \) becomes very large, as in an infinite series, the term \( r^n \) approaches zero if \( |r| < 1 \). This effectively gives the sum of an infinite geometric series:
\( S = \frac{a_1}{1-r} \)
In practical terms, for our series \( \sum_{k=1}^{\infty} \frac{2}{3^{k}} \), we substitute to find that it converges to the value 1, confirming what was previously derived by the limit of partial sums.