Chapter 4: Q. 47 (page 326)

Determine which of the limit of sums in Exercises 47–52 are infinite and which are finite. For each limit of sums that is finite, compute its value.
$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{2} + k + 1}{n}$

Short Answer

Expert verified

The limit of the sum is infinite.

Step by step solution

Step 1. Given information

$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{2} + k + 1}{n}$

Step 2. Find the limit of the sum.

$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{2} + k + 1}{n} = \lim_{n \to \infty} \frac{1}{n} ({\sum k^{2}}_{k = 1}^{n} + \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1) = \lim_{n \to \infty} \frac{1}{n} (\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} + n) = \lim_{n \to \infty} \frac{n}{n} (\frac{(n + 1) (2 n + 1)}{6} + \frac{(n + 1)}{2} + 1) = \lim_{n \to \infty} (\frac{(n + 1) (2 n + 1) + 3 (n + 1) + 6}{6}) = \lim_{n \to \infty} (\frac{2 n^{2} + n + 2 n + 1 + 3 n + 3 + 6}{6}) = \lim_{n \to \infty} (\frac{2 n^{2} + 6 n + 10}{6}) = \infty$

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Determine which of the limit of sums in Exercises 47–52 are infinite and which are finite. For each limit of sums that is finite, compute its value.
$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{2} + k + 1}{n}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Find the limit of the sum.

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Determine which of the limit of sums in Exercises 47–52 are infinite and which are finite. For each limit of sums that is finite, compute its value.limn→∞∑k=1nk2+k+1n

Short Answer

Step by step solution

Step 1. Given information

Step 2. Find the limit of the sum.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Geometry

Decision Maths

Calculus

Discrete Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Determine which of the limit of sums in Exercises 47–52 are infinite and which are finite. For each limit of sums that is finite, compute its value.
$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{2} + k + 1}{n}$