Find the limit of s(n) as n → ∞0. 1 [n(n+1 1) n² 2 9 s(n) :

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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 \( \frac{1}{9} \).
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 \( \frac{1}{9} \).
Expert Solution
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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