Get the most out of your questions with the extensive resources available on IDNLearn.com. Discover detailed and accurate answers to your questions from our knowledgeable and dedicated community members.

What is the value of [tex]\(\lim_{n \rightarrow \infty} S_n\)[/tex] if [tex]\(S_n=\sum_{k=1}^n\left[\frac{24}{n^3} \cdot k^2 + \frac{12}{n^2} \cdot k + \frac{15}{n}\right]\)[/tex]?

Hint: Use the summation formulas from the "Area Under a Curve" lesson.

Answer: [tex]\(\square\)[/tex]


Sagot :

To solve the problem, we need to find the value of [tex]\(\lim _{n \rightarrow \infty} S_n\)[/tex] where [tex]\(S_n\)[/tex] is given by:
[tex]\[S_n = \sum_{k=1}^n \left[ \frac{24}{n^3} k^2 + \frac{12}{n^2} k + \frac{15}{n} \right]\][/tex]

Let's analyze the sum term-by-term.

### Step-by-Step Breakdown:

1. Separate the Sum:
[tex]\[S_n = \sum_{k=1}^n \frac{24}{n^3} k^2 + \sum_{k=1}^n \frac{12}{n^2} k + \sum_{k=1}^n \frac{15}{n}\][/tex]

2. Simplify Each Sum Separately:

#### First Term: [tex]\(\sum_{k=1}^n \frac{24}{n^3} k^2\)[/tex]
[tex]\[\sum_{k=1}^n \frac{24}{n^3} k^2 = \frac{24}{n^3} \sum_{k=1}^n k^2\][/tex]

The formula for the sum of squares of the first [tex]\(n\)[/tex] natural numbers is:
[tex]\[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\][/tex]

Substituting this in:
[tex]\[\frac{24}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{24}{6} \cdot \frac{(n+1)(2n+1)}{n^2} = 4 \cdot \frac{(n+1)(2n+1)}{n^2}\][/tex]

Simplifying further:
[tex]\[\frac{4(n+1)(2n+1)}{n^2} = 4 \cdot \left(\frac{(2n^2 + 3n + 1)}{n^2}\right) = 4 \cdot \left(2 + \frac{3}{n} + \frac{1}{n^2}\right)\][/tex]

As [tex]\(n \rightarrow \infty\)[/tex], [tex]\(\frac{3}{n} \rightarrow 0\)[/tex] and [tex]\(\frac{1}{n^2} \rightarrow 0\)[/tex]:
[tex]\[= 4 \cdot 2 = 8\][/tex]

#### Second Term: [tex]\(\sum_{k=1}^n \frac{12}{n^2} k\)[/tex]
[tex]\[\sum_{k=1}^n \frac{12}{n^2} k = \frac{12}{n^2} \sum_{k=1}^n k\][/tex]

The formula for the sum of the first [tex]\(n\)[/tex] natural numbers is:
[tex]\[\sum_{k=1}^n k = \frac{n(n+1)}{2}\][/tex]

Substituting this in:
[tex]\[\frac{12}{n^2} \cdot \frac{n(n+1)}{2} = \frac{12}{2} \cdot \frac{(n+1)}{n} = 6 \cdot \left(\frac{n+1}{n}\right)\][/tex]

Simplifying further:
[tex]\[6 \cdot \left(1 + \frac{1}{n}\right)\][/tex]

As [tex]\(n \rightarrow \infty\)[/tex], [tex]\(\frac{1}{n} \rightarrow 0\)[/tex]:
[tex]\[= 6 \cdot 1 = 6\][/tex]

#### Third Term: [tex]\(\sum_{k=1}^n \frac{15}{n}\)[/tex]
[tex]\[\sum_{k=1}^n \frac{15}{n} = \frac{15}{n} \sum_{k=1}^n 1 = \frac{15}{n} \cdot n = 15\][/tex]

3. Combine the Results:
Adding up the simplified limits, we get:
[tex]\[ \lim_{n \rightarrow \infty} S_n = 8 + 6 + 15 = 29 \][/tex]

Therefore, the value of [tex]\(\lim _{n \rightarrow \infty} S_n\)[/tex] is:
[tex]\[ \boxed{29} \][/tex]