Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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### Problem Statement **Find the limit of \( s(n) \) as \( n \to \infty \).** ### Function Definition \[ s(n) = \frac{1}{n^2} \left[ \frac{n(n + 1)}{9} \right] \] ### Explanation The equation provided defines the function \( s(n) \) in terms of \( n \). The function includes a term \( n(n + 1) \) which is first divided by 9, and the result is then multiplied by the reciprocal of \( n^2 \). ### Calculating the Limit The goal is to determine the behavior of \( s(n) \) as \( n \) approaches infinity. Consider applying limit laws and algebraic manipulation to evaluate the limit: 1. **Substitute and Simplify:** \[ \text{Inside the brackets: } \frac{n(n + 1)}{9} \approx \frac{n^2}{9} + \frac{n}{9} \] 2. **Calculate the Limit:** \[ s(n) = \frac{1}{n^2} \left( \frac{n^2}{9} + \frac{n}{9} \right) \] \[ s(n) = \frac{1}{9} + \frac{1}{9n} \] 3. **Final Limit:** \[ \lim_{n \to \infty} s(n) = \lim_{n \to \infty} \left( \frac{1}{9} + \frac{1}{9n} \right) = \frac{1}{9} + 0 = \frac{1}{9} \] Thus, the limit of \( s(n) \) as \( n \) approaches infinity is \( \frac{1}{9} \).
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Transcribed Image Text:### Problem Statement **Find the limit of \( s(n) \) as \( n \to \infty \).** ### Function Definition \[ s(n) = \frac{1}{n^2} \left[ \frac{n(n + 1)}{9} \right] \] ### Explanation The equation provided defines the function \( s(n) \) in terms of \( n \). The function includes a term \( n(n + 1) \) which is first divided by 9, and the result is then multiplied by the reciprocal of \( n^2 \). ### Calculating the Limit The goal is to determine the behavior of \( s(n) \) as \( n \) approaches infinity. Consider applying limit laws and algebraic manipulation to evaluate the limit: 1. **Substitute and Simplify:** \[ \text{Inside the brackets: } \frac{n(n + 1)}{9} \approx \frac{n^2}{9} + \frac{n}{9} \] 2. **Calculate the Limit:** \[ s(n) = \frac{1}{n^2} \left( \frac{n^2}{9} + \frac{n}{9} \right) \] \[ s(n) = \frac{1}{9} + \frac{1}{9n} \] 3. **Final Limit:** \[ \lim_{n \to \infty} s(n) = \lim_{n \to \infty} \left( \frac{1}{9} + \frac{1}{9n} \right) = \frac{1}{9} + 0 = \frac{1}{9} \] Thus, the limit of \( s(n) \) as \( n \) approaches infinity is \( \frac{1}{9} \).
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Step 1: Given

S left parenthesis n right parenthesis equals 1 over n squared open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 9 end fraction close square brackets

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