Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.1 - Page 243: 50

Answer

$\frac{1}{2!}$ + $\frac{2}{3!}$ + $\frac{3}{4!}$ + ... + $\frac{n}{(n+1)!}$ $\sum_{k=0}^N \frac{k+1}{(k+2)!}$

Work Step by Step

The sequence begins from 0 until n . When k=0, the term (i.e, that particular element of the sequence) is evaluated to $\frac{0+1}{(0+2)!}$ . This is why the term is $\frac{1}{2!}$ When k=1, the term (i.e, that particular element of the sequence) is evaluated to $\frac{1+1}{(1+2)!}$ . This is why the term is $\frac{2}{3!}$ When k = 2, the term (i.e, that particular element of the sequence) is evaluated to $\frac{2+1}{(2+2)!}$. This is why the term is $\frac{3}{4!}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.