Chapter 4: Q. 49 (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
$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1}$

Short Answer

Expert verified

The limit of the sum is finite and it is $\frac{1}{3}$ .

Step by step solution

Step 1. Given information

$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1}$

Step 2. Find the limit of the sum.

$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1} = \lim_{n \to \infty} \frac{1}{n^{3} - 1} \sum_{k = 1}^{n} {(k + 1)}^{2} = \lim_{n \to \infty} \frac{1}{n^{3} - 1} \sum_{k = 1}^{n} (k^{2} + 2 k + 1) = \lim_{n \to \infty} \frac{1}{n^{3} - 1} (\sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} 2 k + \sum_{k = 1}^{n} 1) = \lim_{n \to \infty} \frac{1}{n^{3} - 1} (\sum_{k = 1}^{n} k^{2} + 2 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1) = \lim_{n \to \infty} \frac{1}{n^{3} - 1} (\frac{n (n + 1) (2 n + 1)}{6} + 2 \frac{n (n + 1)}{2} + n) = \lim_{n \to \infty} \frac{1}{n^{3} - 1} (\frac{2 n^{3} + 3 n^{2} + n}{6} + n^{2} + n + n) = \lim_{n \to \infty} \frac{1}{n^{3} - 1} (\frac{2 n^{3} + 3 n^{2} + n + 6 n^{2} + 12 n}{6}) = \lim_{n \to \infty} (\frac{2 n^{3} + 9 n^{2} + 13 n}{6 n^{3} - 6}) = \lim_{n \to \infty} \frac{n^{3} (2 + \frac{9}{n} + \frac{13}{n^{2}})}{n^{3} (6 - \frac{6}{n^{3}})} = \lim_{n \to \infty} \frac{(2 + \frac{9}{n} + \frac{13}{n^{2}})}{(6 - \frac{6}{n^{3}})} = \frac{2}{6} = \frac{1}{3}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

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
$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1}$

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

Decision Maths

Theoretical and Mathematical Physics

Applied Mathematics

Pure Maths

Mechanics Maths

Logic and Functions

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 valuelimn→∞∑k=1nk+12n3-1

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

Decision Maths

Theoretical and Mathematical Physics

Applied Mathematics

Pure Maths

Mechanics Maths

Logic and Functions

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
$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1}$