1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is it an Alternating Series?

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Is this an alternating series : (-1)^n / ( arctan n ) ^ (n)

    2. Relevant equations

    I know that the alternating series should have (-1) to any power, but also the signs should be - , + , - , + ..... or the opposit,

    3. The attempt at a solution

    I am ok with this I know that for testing Con. Or Div. of this we should use the root test

    But this series. But the problem is " Is this seris alternating or not"

    Becasue if not it will be treated normally with a root test and it will Con.

    For me I think that it is not "Alternating "

    What do you think ?
     
  2. jcsd
  3. Apr 11, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Does 1/arctan(n)^n decrease to zero? That's what they want to you check.
     
  4. Apr 11, 2009 #3
    mmm, do you mean the limit, if so no. It goes to inifity.
     
  5. Apr 11, 2009 #4
    So, this is NOT an alternating sries, right ????
     
  6. Apr 11, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No. 1/arctan(n)^n doesn't go to infinity. Tell me why you think it does.
     
  7. Apr 11, 2009 #6
    ohhh, Ya , I remembred it goes to Pie/ 2
    Right ?
     
  8. Apr 12, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    arctan(n) goes to pi/2. 1/arctan(n)^n doesn't.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is it an Alternating Series?
  1. Alternating series (Replies: 3)

  2. Alternating Series (Replies: 1)

  3. Alternating Series (Replies: 2)

  4. Alternating Series (Replies: 5)

  5. Alternating series (Replies: 2)

Loading...