Question

Calculus. Find the sum

Answer 1

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

Answer 2

Try converting your fraction into a partial fraction decomposition. Then, you'll notice you have a subtraction of 2 fractions.

Then, start plugging in a few terms... starts with n=3, n=4, n=5, n=6, n=7. Write those down and notice if you see any cancellations ;) You'll notice it's a telescoping series!

Answer 3

I have tried this. I cant seem to get the correct sum

Answer 4

Hm. What did you get as your partial fraction decomposition

Answer 5

\[\frac{ 4 }{ n+1 }-\frac{ 4 }{ n-1 }\]

Answer 6

i got a= 4 and b = -4

Answer 7

i keep getting 4/2= 2

Answer 8

hmm it's not \[-\frac{ 4 }{ n+1 }+\frac{ 4 }{ n-1 }\]?

Answer 9

i typed wrong. i did the n-1 first. does it matter?

Answer 10

euh well I would think it would affect the result

Answer 11

should i do the n+1 with A and the -4?

Answer 12

that would make the sum -1?

Answer 13

yes

Answer 14

euh no not the sum

Answer 15

\[\frac{4}{2}-\frac{4}{4}+\frac{4}{3}-\frac{4}{5}+\frac{4}{4}-\frac{4}{6}+\frac{4}{5}-\frac{4}{7}+\cdots\]

Answer 16

notice which terms won't cancel out near the beginning of the sum

Answer 17

4/2 and 4/3

Answer 18

yes :)

Answer 19

just add those and that is the sum?

Answer 20

bingo =]

Answer 21

so, you did the n-1 first?

Answer 22

yeah I did.

Answer 23

thanks

Answer 24

no problem!