## anonymous 5 years ago 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..)

1. anonymous

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

2. anonymous

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

3. anonymous

yes haha

4. anonymous

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.

5. anonymous

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

6. anonymous

yup got that, i'm having trouble simplifying that

7. anonymous

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

8. anonymous

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

9. anonymous

further, taking the limit of the already established reciprocal

10. anonymous

multiplaction step

11. anonymous

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

12. anonymous

finding the limit

13. anonymous

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

14. anonymous

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

15. anonymous

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.

16. anonymous

Let me look in my notes real quick

17. anonymous

ok :)

18. anonymous

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

19. anonymous

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

20. anonymous

The factorials cancel each other in a way.

21. anonymous

..and you'll keep the constants right?

22. anonymous

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.

23. anonymous

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

24. anonymous

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

25. anonymous

I believe so

26. anonymous

wonderful :)

27. anonymous