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Q. 52

Expert-verified

Calculus

Found in: Page 326

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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
$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{3}}{n^{4} + n + 1}$

The limit of the sum is finite and it is equal to $\frac{1}{4}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

Given :

\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{3}}{n^{4} + n + 1}

Step 2. Find limit of the sum.

$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{3}}{n^{4} + n + 1} = \lim_{n \to \infty} \frac{1}{n^{4} + n + 1} \sum_{k = 1}^{n} k^{3} = \lim_{n \to \infty} \frac{1}{n^{4} + n + 1} {(\frac{n (n + 1)}{2})}^{2} = \lim_{n \to \infty} \frac{1}{n^{4} + n + 1} (\frac{n^{2} {(n + 1)}^{2}}{4}) = \lim_{n \to \infty} \frac{1}{n^{4} + n + 1} (\frac{n^{2} (n^{2} + 2 n + 1)}{4}) = \lim_{n \to \infty} \frac{n^{4} (1 + \frac{2}{n} + \frac{1}{n^{2}})}{4 n^{4} (1 + \frac{1}{n^{3}} + \frac{1}{n^{4}})} = \lim_{n \to \infty} \frac{(1 + \frac{2}{n} + \frac{1}{n^{2}})}{4 (1 + \frac{1}{n^{3}} + \frac{1}{n^{4}})} = \frac{1 + 0 + 0}{4 (1 + 0 + 0)} = \frac{1}{4}$

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