Problem 9 - Calculus: One and Several Variables

Question

Find the sum of the series. $$\sum_{k=0}^{\infty} \frac{2^{k+3}}{3^{k}}$$

Step-by-Step Solution

Answer

The sum of the series is 24.

Step 1: Identify the first term(a) and the common ratio(r)

From the given series, the first term is 8 (which is the k = 0 term in the series or \(2^{0+3}\)) and the common ratio is \( \frac{2}{3} \), which is the ratio of each successive term (2^{k+1}/3^k) to its preceding term (2^k/3^k).

Step 2: Substituting into the formula for sum of an infinite geometric series

The sum of an infinite geometric series is given by the formula \(S = \frac{a}{1-r}\). So subtitute a = 8 and r = \( \frac{2}{3} \) into the formula. Therefore, \(S = \frac{8}{1 - \frac{2}{3}}\)

Step 3: Simplify the expression

To simplify \(S = \frac{8}{1 - \frac{2}{3}}\), we can rewrite the denominator as a fraction with a common denominator, simplify it and then multiply 8 by the reciprocal.This gives us, \(S = 8 * \frac{3}{1}\)

Step 4: Compute the final total

After simplifying further, we get \(S = 8 * 3 = 24\).

Key Concepts

Sum of Series

An infinite geometric series is a series where the terms are continuously added together without an end. These series have a defined rule that we can use to calculate their sum when the absolute value of the common ratio is less than one. The formula to find the sum, denoted as \( S \), of such a series is:

• \( S = \frac{a}{1-r} \)

Here,

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

To solve our original problem, the series given was \( \sum_{k=0}^{\infty} \frac{2^{k+3}}{3^{k}} \). We identified that the first term \( a \) is 8 and the common ratio \( r \) is \( \frac{2}{3} \).
Substituting these values into the formula, we calculate the sum:

• \( S = \frac{8}{1 - \frac{2}{3}} \)
• \( S = \frac{8}{\frac{1}{3}} \)
• \( S = 8 \times 3 \)
• So, \( S = 24 \)

This result tells us that despite the infinite number of terms, the series sums up to a finite value of 24.

First Term of Series

For an infinite geometric series, identifying the first term accurately is critical because it acts as the starting point of the sequence. In any series, the first term \( a \) is generally represented when the index \( k \) is zero.
The original series was:

• \( \sum_{k=0}^{\infty} \frac{2^{k+3}}{3^{k}} \).

To find the first term for this specific series, substitute \( k = 0 \) in the term formula:

• \( \frac{2^{0+3}}{3^{0}} = \frac{2^3}{1} = 8 \)

Thus, the first term \( a \) is 8, which gives us a solid base for using our formula for the sum of the series.

Common Ratio

The common ratio in a geometric series is the factor by which each term is multiplied to get the next term. In order to determine the common ratio \( r \) in an infinite geometric series, you need to consider any two consecutive terms.
For our given series, \( \frac{2^{k+3}}{3^{k}} \), we compared:

• The first term: \( \frac{2^{0+3}}{3^{0}} = 8 \)
• And the second term: \( \frac{2^{1+3}}{3^{1}} = \frac{16}{3} \)

To find the common ratio, divide the second term by the first term:

• \( r = \frac{\frac{16}{3}}{8} = \frac{16}{3} \times \frac{1}{8} = \frac{2}{3} \)

Ensuring that this ratio is less than one is crucial because only then can the infinite series converge to a finite sum. In this scenario, \( r \) being \( \frac{2}{3} \) satisfies this condition, allowing us to use the finite sum formula.