Question

Which of the following converges conditionally?

Answer 1

Multiple choice also

Answer 2

Okay, take your time.

Answer 3

a. summation (-1)^n (1/3)^(n-1)
b. summation 2 infinity (-1)^n (1/sqr )n))
c. summation 2 to infintity (-1)^n (1/n^2)
d.) 1-1.1+1.21-1.331+.....
e.) 1/(1x2)-1/(2x3)+1/(3x4)-1/(4x5)....

Answer 4

are the summations from 0 to infinity or 1 to infinity (just curious)

Answer 5

1 to infinity for a

Answer 6

okay, well for conditional convergence you can do the alternating series test. Do you remember what that says?

Answer 7

if Bn+1 < bn
lim bn = 0
it is convergant right?

Answer 8

exactly, so a you have bn+1<bn or (1/3)^n<(1/3)^(n-1) which is true for all n>1. And but the limit as bn->inf =/= 0 so that one diverges by the limit test for divergence

Answer 9

conditionally is when one part converges...but not all right? (sorry this is review for me)

Answer 10

For b, you get bn+1<bn 1/sqrt(n+1)<1/sqrt(n)-> sqrt(n)<sqrt(n+1) which is true for all n>2 and the limit n->inf of 1/sqrt(n) DOES tend to zero so the series converges conditionally.

Answer 11

Conditional is when it converges ONLY if there is a negative (alternation) absolute means it doesn't matter if there is alternation or not.

Answer 12

Oooo okay I see

Answer 13

Thank you :) very much!

Answer 14

a perfect example is the sum 1 to infinity 1/n diverges but the sum 1 to infinity of (-1)^n/n CONVERGES. No problem :)

Answer 15

That's all I have thank you!

Answer 16

No problem :) You can get me anytime you need help!

Answer 17

how? lol

Answer 18

I'm not sure D: My facebook is linked you can probably just click that and send me a message xD

Answer 19

Oooo sweet thank you :D

Answer 20

No problem :) If nothing else you can go to the mathematics page itself and it should show all the people online, just ctrl+f my name xP

Answer 21

alright! >w<

Answer 22

See ya :D