1. Limited time only! Sign up for a free 30min personal 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!

Question about root test for series

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Can you if given a sigma notation for an infinite series

    Ʃ 2^n/(4^n+1)
    rewrite as,

    Ʃ 2^n/(4^n+1^n)
    then doing the root test can you

    lim n→∞ n^√abs((2/(4+1))^n)

    which equals 2/5, and
    2/5<1, therefore can i conclude that the series above converges?

    Sorry if this is in the wrong section, i guess its more of an algebra question.
     
  2. jcsd
  3. Apr 6, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    No this is 100% wrong to do.

    Your series is :

    ##\sum \frac{2^n}{4^{n+1}}##

    Which is definitely NOT equal to :

    ##\sum \frac{2^n}{4^{n} + 1^{n}}##

    Why don't you try the ratio test, it will work perfectly for this.
     
  4. Apr 6, 2013 #3
    sorry i didnt put enough parentheses my series is 2^n/((4^n)+1)
     
  5. Apr 6, 2013 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member


    OP was correct !

    Actually, Ʃ 2^n/(4^n+1) is [itex]\displaystyle \ \sum\frac{2^n}{4^n+1}\ .[/itex]
     
  6. Apr 6, 2013 #5

    Zondrina

    User Avatar
    Homework Helper

    Sorry, the lack of parenthesis confused me.
     
  7. Apr 7, 2013 #6
    So can that 1 be represented by a 1^n? Because no matter what n is Given n>=0 that 1^n = 1.?
     
  8. Apr 7, 2013 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes. That's fine, but I don't see why that helps.

    You do realize that [itex]\displaystyle \ 4^n+1^n\ne(4+1)^n\,,\ [/itex] don't you ?
     
  9. Apr 7, 2013 #8
    I thought that it looked funky when I did it. Thank you though
     
  10. Apr 7, 2013 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    So, do you have a solution ?
     
  11. Apr 7, 2013 #10
    Using the ratio test because there are constants raised to powers i get..
    (2^(n+1)/(4^(n+1)+1))*((4^(n)+1)/2^n) then using some algerbra
    (4^(n)+1)/(4*4^(n)+1) and some more algerbra....
    gives me 1/4 which is <1 therefore the series converges?
     
  12. Apr 7, 2013 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Are you using a limit without stating that you are?

    [itex]\displaystyle \frac{4^n+1}{4^{n+1}+1}\ne \frac{1}{4}\ .[/itex]

    Also, what happened to the 2 ?
     
  13. Apr 7, 2013 #12
    yes i am using a limit and i forgot about the 2, i guess it converges to 1/2 then? i had forgotten that the ratio test needed a limit because i wasnt in class when we learned it.
     
  14. Apr 7, 2013 #13

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Anyway, the series converges.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Question about root test for series
Loading...