Question

7. [11 points] Find the sum of the series ( sum_{n=1}^{infty} frac{6}{n(n+3)} ) by writing it as a telescoping sum, using the fact: [ frac{6}{n(n+3)}=frac{2}{n}-frac{2}{n+3} ]

          7. [11 points] Find the sum of the series ( sum_{n=1}^{infty} frac{6}{n(n+3)} ) by writing it as a telescoping sum, using the fact:
[
frac{6}{n(n+3)}=frac{2}{n}-frac{2}{n+3}
]

$7. [11 points] Find the sum of the series ( sumn=1^infty frac6n(n+3) ) by writing it as a telescoping sum, using the fact: [ frac6n(n+3)=frac2n-frac2n+3 ]$

Added by Jay K.

Algebra 2

Holliday, Luchin, Cuevas, Carter Marks,… 1st Edition

Chapter 11

Instant Answer

Solved by Expert Ekaveera Kumar

05/08/2022

Step 1

Step 1: Rewrite the given summation as $ \sum_{n=1}^{\infty} \frac{6}{n(n+3)} $ as $ \sum_{n=1}^{\infty} \left( \frac{3}{n} - \frac{3}{n+3} \right) $. Show more…

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