11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from k = 0 to ∞)k² · 1.001⁻ᵏ

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Identify the given series: \( \sum_{k=0}^{\infty} k^{2} \cdot 1.001^{-k} \). Notice that the general term is \( a_k = k^{2} \cdot (1.001)^{-k} \).

Rewrite the term to see the exponential part clearly: \( a_k = k^{2} \cdot \left( \frac{1}{1.001} \right)^{k} \). Since \( \frac{1}{1.001} < 1 \), this suggests the terms involve a polynomial factor times an exponential decay.

Consider applying the Ratio Test, which is useful for series involving factorials, exponentials, or powers. The Ratio Test states to compute \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).

Calculate \( \frac{a_{k+1}}{a_k} = \frac{(k+1)^2 \cdot (1.001)^{-(k+1)}}{k^2 \cdot (1.001)^{-k}} = \frac{(k+1)^2}{k^2} \cdot \frac{1}{1.001} \).

Evaluate the limit \( L = \lim_{k \to \infty} \frac{(k+1)^2}{k^2} \cdot \frac{1}{1.001} = \left( \lim_{k \to \infty} \frac{(k+1)^2}{k^2} \right) \cdot \frac{1}{1.001} = 1 \cdot \frac{1}{1.001} < 1 \). Since \( L < 1 \), the Ratio Test confirms that the series converges.

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Geometric Series and Its Convergence

A geometric series has terms of the form ar^k, where r is the common ratio. It converges if |r| < 1 and diverges otherwise. Understanding this helps analyze series with exponential terms like 1.001⁻ᵏ, which can be rewritten as (1/1.001)^k, a geometric term with ratio less than 1.

Comparison and Limit Comparison Tests

These tests compare a given series to a known benchmark series to determine convergence. The limit comparison test is useful when terms involve polynomials and exponentials, helping to confirm if the series behaves like a convergent geometric series despite polynomial factors like k².

Effect of Polynomial Factors on Series Convergence

Polynomial terms like k² grow slower than exponential decay terms. When combined with a geometric term with ratio less than 1, the exponential decay dominates, often ensuring convergence. Recognizing this balance is key to justifying convergence in series with mixed polynomial and exponential terms.

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.∑ (from k = 0 to ∞)k² · 1.001⁻ᵏ