Question

Find the sum of the convergent series \[\sum_{n=1}^{\infty} 1/n(n+3)\]

Answer 1

Is it 11/6?

Answer 2

11/18

Answer 3

ok 11/18

Answer 4

now let me check

Answer 5

do you know how to write 1/n(n+3) in polynomial form?

Answer 6

1/n^2+3n?

Answer 7

no i mean it can be written as A/n - B/(n+3)

Answer 8

oh partial fractions

Answer 9

if you multiply the first part by n(+3) and the second by (n) you get A*(n+3)-B*n=1

Answer 10

yeah partial fractions

Answer 11

here A=1/3 and B=1/3

Answer 12

B= - 1/3?

Answer 13

forgot the negative on b

Answer 14

And sum of ne from 1 to inf. (1/3 / n) - (1/3 / (n-3)

Answer 15

I fyou start inserting numbers from 1; you get:
1/3[ 1+1/2+1/3+1+4+......) - 1/3( 1/4+1/5+1/6+.....)

Answer 16

you see they cancel from 1/4 and only 1/3(1+1/2+1/3) remains

Answer 17

and that is 11/18

Answer 18

okay thank you!

Answer 19

telescope it until the terms can be cxl, then take the limit of the terms that are not numbers, then sum the numbers to get the final answer