Q. 49 Determine which of the limit of ... [FREE SOLUTION]

91影视

Chapter 4: Q. 49 (page 326)

Determine which of the limit of sums in Exercises 47鈥�52 are infinite and which are finite. For each limit of sums that is finite, compute its value
$\lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1}$

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

$\lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1}$

Step 2. Find the limit of the sum.

$\lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1} = \lim_{n 鈫� 鈭�} \frac{1}{n^{3} - 1} {鈭�}_{k = 1}^{n} {(k + 1)}^{2} = \lim_{n 鈫� 鈭�} \frac{1}{n^{3} - 1} {鈭�}_{k = 1}^{n} (k^{2} + 2 k + 1) = \lim_{n 鈫� 鈭�} \frac{1}{n^{3} - 1} ({鈭�}_{k = 1}^{n} k^{2} + {鈭�}_{k = 1}^{n} 2 k + {鈭�}_{k = 1}^{n} 1) = \lim_{n 鈫� 鈭�} \frac{1}{n^{3} - 1} ({鈭�}_{k = 1}^{n} k^{2} + 2 {鈭�}_{k = 1}^{n} k + {鈭�}_{k = 1}^{n} 1) = \lim_{n 鈫� 鈭�} \frac{1}{n^{3} - 1} (\frac{n (n + 1) (2 n + 1)}{6} + 2 \frac{n (n + 1)}{2} + n) = \lim_{n 鈫� 鈭�} \frac{1}{n^{3} - 1} (\frac{2 n^{3} + 3 n^{2} + n}{6} + n^{2} + n + n) = \lim_{n 鈫� 鈭�} \frac{1}{n^{3} - 1} (\frac{2 n^{3} + 3 n^{2} + n + 6 n^{2} + 12 n}{6}) = \lim_{n 鈫� 鈭�} (\frac{2 n^{3} + 9 n^{2} + 13 n}{6 n^{3} - 6}) = \lim_{n 鈫� 鈭�} \frac{n^{3} (2 + \frac{9}{n} + \frac{13}{n^{2}})}{n^{3} (6 - \frac{6}{n^{3}})} = \lim_{n 鈫� 鈭�} \frac{(2 + \frac{9}{n} + \frac{13}{n^{2}})}{(6 - \frac{6}{n^{3}})} = \frac{2}{6} = \frac{1}{3}$

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Determine which of the limit of sums in Exercises 47鈥�52 are infinite and which are finite. For each limit of sums that is finite, compute its value
$\lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1}$

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

Geometry

Logic and Functions

Discrete Mathematics

Theoretical and Mathematical Physics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

91影视

Determine which of the limit of sums in Exercises 47鈥�52 are infinite and which are finite. For each limit of sums that is finite, compute its valuelimn鈫�鈭�鈭�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

Geometry

Logic and Functions

Discrete Mathematics

Theoretical and Mathematical Physics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Determine which of the limit of sums in Exercises 47鈥�52 are infinite and which are finite. For each limit of sums that is finite, compute its value
$\lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} \frac{{(k + 1)}^{2}}{n^{3} - 1}$