Answer
3
Work Step by Step
We see, from the question 20 of this exercise, that
$\Sigma^{n=\infty}_{n=1} \frac{1}{n(n+2)}=\frac {1}{2} \times\frac{3}{2} = \frac{3}{4}$
Therefore, $\Sigma^{n=\infty}_{n=1} \frac{4}{n(n+2)}= 4\times\frac{3}{4}=3$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 9 - Infinite Series - 9.2 Exercises - Page 601: 27
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
