Question

Write down the first three partial sums for the infinite series. I will write the series with the equation box.

Answer 1

Give me a sec 2 put teh equation in

Answer 2

\[\sum_{n=1}^{\infty} [(-1)^n-1/ [n(n+1)]\]

Answer 3

CORRECTION. The numerator shud be (-1) to the (n-1) power

Answer 4

and the denominator is same: n(n+1)

Answer 5

\[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{ n(n+1)} \]

Answer 6

for the first term, replace n with 1

Answer 7

-1^(0) = 1
1(1+2) = 2

Answer 8

so 1/2

Answer 9

*1(1+1) = 2

Answer 10

wat does the star mean

Answer 11

typo

Answer 12

the asnwer is 1/2 though not 2

Answer 13

Yes, Look up a few boxes. I mean to type
-1^(0) = 1
1(1+1) = 2

where the first line is the top of the fraction, and the second line is the denominator

Answer 14

oh ok i see

Answer 15

try the next term, with n=2

Answer 16

so i just write S sub 2 = -1/6

Answer 17

-1^(2-1)= -1^1= -1
2(2+1)=2*3= 6
so -1/6 Yes

Answer 18

NVR MIND teh 1/6

Answer 19

You are just interpreting short hand, because math people are lazy and don't want to write down all the terms...

Answer 20

so S sub 3 is = 1/12

Answer 21

but to find teh partial sum dont u have tofind a pattern and then write it as S sub n = sum equation

Answer 22

Yes, we should be adding them up. Did I forget to do that?
so 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3 for the 2nd partial sum

Answer 23

and the 3rd should be
1/3+1/12 = 4/12+1/12 = 5/12

Answer 24

i see =) tahnks. can i ask u to help me with another question about convergence/divergence?

Answer 25

i will post it

Answer 26

ok

Answer 27

shud i put it here or post it on teh side

Answer 28

new post. Otherwise, these things get too long.

Answer 29

ok just give me a sec