Question

telescopic series

Answer 1

\[\sum_{n=3}^{\infty}\frac{ 8 }{ n^2-1 }\]

Answer 2

Partial fractions:
\[\frac{8}{n^2-1}=\frac{8}{(n-1)(n+1)}=4\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\]

Answer 3

what do we do after this?

Answer 4

Check to see if a pattern emerges.
\[\sum_{n=3}^\infty4\left(\frac{1}{n-1}-\frac{1}{n+1}\right)=4\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\cdots\right)\]
Which terms disappear? Which terms remain?

Answer 5

when you do this, how do you know how far to go?

Answer 6

from what i can tell, the 1/4 's disappear, but how far should we go to see what else disappears?

Answer 7

Write out a few more terms:
\[\underbrace{\frac{1}{2}\color{red}{-\frac{1}{4}}}_{\text{first term}}+\underbrace{\frac{1}{3}\color{blue}{-\frac{1}{5}}}_{\text{second}}+\underbrace{\color{red}{\frac{1}{4}}-\color{green}{\frac{1}{6}}}_{\text{third}}+\underbrace{\color{blue}{\frac{1}{5}}-\color{pink}{\frac{1}{7}}}_{\text{fourth}}+\cdots\]
The \(\dfrac{1}{4}\) terms are the first to disappear, then \(\dfrac{1}{5}\), and so on, so you're left with just \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\).

Answer 8

im still working on this problem lol