27–37. Evaluating series Evaluate the following infinite series or state that the series diverges.
∑ (from k = 1 to ∞)2ᵏ / 3ᵏ⁺²

Verified step by step guidance

1. Identify the given series: \( \sum_{k=1}^{\infty} \frac{2^k}{3^{k+2}} \). Notice that the series involves terms with exponents depending on \(k\).
2. Rewrite the general term to separate the powers of 3: \( \frac{2^k}{3^{k+2}} = \frac{2^k}{3^k \cdot 3^2} = \frac{1}{3^2} \cdot \frac{2^k}{3^k} = \frac{1}{9} \cdot \left( \frac{2}{3} \right)^k \).
3. Recognize that the series is a geometric series with the first term \( a = \frac{1}{9} \cdot \left( \frac{2}{3} \right)^1 = \frac{2}{27} \) and common ratio \( r = \frac{2}{3} \).
4. Check the convergence of the geometric series by verifying if \( |r| < 1 \). Since \( \frac{2}{3} < 1 \), the series converges.
5. Use the formula for the sum of an infinite geometric series starting at \( k=1 \): \[ S = a \cdot \frac{1}{1 - r} \], where \( a \) is the first term and \( r \) is the common ratio. Substitute the values to express the sum.

Key Concepts

Infinite Geometric Series

An infinite geometric series is a sum of terms where each term is found by multiplying the previous term by a constant ratio. It converges if the absolute value of the ratio is less than 1, and its sum can be calculated using the formula S = a / (1 - r), where a is the first term and r is the common ratio.

Convergence and Divergence of Series

A series converges if the sum of its infinite terms approaches a finite limit; otherwise, it diverges. For geometric series, convergence depends on the common ratio's magnitude. Understanding convergence is essential to determine whether the series sum exists or not.

Manipulating Series Terms

Simplifying the general term of a series often involves algebraic manipulation, such as factoring exponents or rewriting terms to identify the first term and common ratio. This step is crucial to apply known formulas and test for convergence effectively.