Question

determine if the summation to infinity of the alternating sum is absolutely convergent, conditionally convergent or divergent: (-1)^(k+1)/k! (having problems simplifying the absolute convergence test, for some reason i was never taught how to simplify the factorial when it was in a fraction..)

Answer 1

Never heard it called the absolute convergence test. Are you talking about the ratio test?

Answer 2

\[\lim_{n \rightarrow \infty} \left| a(n+1)/a(n) \right|\] ?

Answer 3

yes haha

Answer 4

Ah ok. Well what your going to do is you write your function out, replacing "n" with "n+1". Then your going to divide it by your function written with just "n" for your variable.

Answer 5

But you can bypass that step by multiplying by the reciprocal once you get the hang of it.

Answer 6

yup got that, i'm having trouble simplifying that

Answer 7

Your k! when your replacing "k" with "k+1" will end up being "(k+1)!"

Answer 8

Is that what you meant or as you progress farther into the problem when you have k! over (k+1)!

Answer 9

further, taking the limit of the already established reciprocal

Answer 10

multiplaction step

Answer 11

ok. And what exact thing are you having trouble multiplying?

Answer 12

finding the limit

Answer 13

right now i have 1^(k+1)+1/(k+1)! * k!/1^(k+1)

Answer 14

so whats next? how do i find that limit/simplifiy that fraction

Answer 15

Ok. there is a property that comes into play when you have n!/(n+1)!. I'm trying to think of how to show it. It reduces it down.

Answer 16

Let me look in my notes real quick

Answer 17

ok :)

Answer 18

cause i'll also need that rule to test the original function for convergence

Answer 19

Ok here is what happens. Lets say you have n! over (n+1)!. Your result will be (n+1) in the denominator (no factorial)

Answer 20

The factorials cancel each other in a way.

Answer 21

..and you'll keep the constants right?

Answer 22

Well the only thing affected in what I just told you is your k! and (k+1)! The rest of the problem reduces just like typical multiplication & cancelling.

Answer 23

So the first term in what you have right now has a (k+1) in the denominator instead of (k+1)!

Answer 24

so i'll end up taking the limit of 1/k+1 ?

Answer 25

I believe so

Answer 26

wonderful :)

Answer 27

thanks for your help! :)

Answer 28

Your very welcome