Jump to content

  • Log in with Facebook Log in with Twitter Log In with Google Sign In
  • Create Account
Welcome to IB Survival
Register now to gain access to all of our features. Once registered and logged in, you will be able to create topics, post replies to existing threads, give reputation to your fellow members, get your own private messenger, post status updates, manage your profile and so much more. If you already have an account, login here - otherwise create an account for free today!
Please browse through the links below for more information. How to download files | How to become VIP | How to contribute files | Questions

Need help on ratio test.

- - - - -

  • Please log in to reply
7 replies to this topic

#1
Excluded

Excluded
  • Members
  • Unknown
  • 6 posts
  • Local time: 03:46 AM
  • Exams: May 2012
  • Canada

Current mood: None chosen
My teacher haven't taught the class how to use the ratio test. I want to apply the ratio test for my math portfolio and unfortunately it's due on Monday, therefore I cannot ask my teacher how to use the ratio test.
I would like to know if L = 0.
Attached File  l.png   2.18K   4 downloads

So far I got up to here:
Attached File  2.png   1.35K   3 downloads

I have no clue what I should do from here.

Thank you in advance.

Advert



#2
kiwi.at.heart

kiwi.at.heart

    The Confused One

  • VIP
  • Fantastic
  • 251 posts
  • Local time: 07:46 PM
  • Exams: Nov 2011
  • Australia

Current mood: Insomnious
At this point you would sub in for n as n-->
As n-->∞, n+1-->∞ so L-->0

#3
rFumachi

rFumachi
  • Members
  • Good
  • 99 posts
  • Local time: 04:46 AM
  • Exams: Nov 2011
  • Peru

Current mood: Hung Over
Attached File  Capture.GIF   6.65K   4 downloads
Why do you want to know if L = 0? It is enough to know that L is less than 1, to know that the series converges absolutely.
What portfolio are you doing?

#4
Excluded

Excluded
  • Members
  • Unknown
  • 6 posts
  • Local time: 03:46 AM
  • Exams: May 2012
  • Canada

Current mood: None chosen

View PostrFumachi, on Nov 26, 2011 - 23:12, said:

Attachment Capture.GIF
Why do you want to know if L = 0? It is enough to know that L is less than 1, to know that the series converges absolutely.
What portfolio are you doing?

I'm working on infinite summation

[Edit]
Another question... How is |1/(n+1)| = 0?
Thanks in advance

Edited by Excluded, Nov 26, 2011 - 23:30.


#5
The Economist

The Economist

    Economics Geek

  • Global Moderator
  • Incredible
  • 477 posts
  • Local time: 12:46 PM
  • Exams: May 2011
  • Greece

Current mood: Cynical

View PostExcluded, on Nov 26, 2011 - 23:19, said:

View PostrFumachi, on Nov 26, 2011 - 23:12, said:

Attachment Capture.GIF
Why do you want to know if L = 0? It is enough to know that L is less than 1, to know that the series converges absolutely.
What portfolio are you doing?

I'm working on infinite summation

[Edit]
Another question... How is |1/(n+1)| = 0?
Thanks in advance

The lim of |1/(n+1)| as n tends to infinity is zero.

#6
rFumachi

rFumachi
  • Members
  • Good
  • 99 posts
  • Local time: 04:46 AM
  • Exams: Nov 2011
  • Peru

Current mood: Hung Over
When you divide 1 by a number that is increasingly larger (like infinity + 1), then the answer will be smaller, tending to zero. If you have any doubt graph it :D

Edited by rFumachi, Nov 26, 2011 - 23:55.


#7
Excluded

Excluded
  • Members
  • Unknown
  • 6 posts
  • Local time: 03:46 AM
  • Exams: May 2012
  • Canada

Current mood: None chosen
oh right, thanks~~ :D i was letting n= 0 thus 1/(0)+1 = 1 Although this is true, I just needed to let n continue to infinity.

#8
rFumachi

rFumachi
  • Members
  • Good
  • 99 posts
  • Local time: 04:46 AM
  • Exams: Nov 2011
  • Peru

Current mood: Hung Over

View PostExcluded, on Nov 27, 2011 - 00:03, said:

oh right, thanks~~ :D i was letting n= 0 thus 1/(0)+1 = 1 Although this is true, I just needed to let n continue to infinity.
You're welcome :) Self-studying series isn't that easy XD






Log In or Register
Register or login to IB Survival to hide some of the ads and gain access to additional features