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Convergence of a sequence

  1. Apr 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine if the sequence is convergent or divergent
    {[tex]\sqrt[n]{3^n+5^n}[/tex]}


    2. Relevant equations



    3. The attempt at a solution
    I know I need to take the limit to find if it converges or diverges. But I'm not really sure what I need to do to it to take the limit.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 20, 2008 #2

    Dick

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    Take the log of the sequence. Write 3^n+5^n=5^n(1+(3/5)^n). Does that help?
     
  4. Apr 21, 2008 #3
    no need to that even, use the root test, and from there you can tell
     
  5. Apr 21, 2008 #4
    I thought you could only apply the root test to a series, not a sequence, or does it not matter?
     
  6. Apr 21, 2008 #5

    Dick

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    Right. The root test is not helpful for sequences.
     
  7. Apr 21, 2008 #6
    so after taking the log I came up with 5(1+(3/5)) = 8. Which I don't think is the right answer, did I do something wrong?
     
  8. Apr 21, 2008 #7

    Dick

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    Yes, quite a bit. Review your rules of logs. Try again. log([5^n*(1+3^n/5^2)]^(1/n)). What's your first step?
     
  9. Apr 21, 2008 #8
    oh ok....

    so I bring the 1/n in front of the ln(5^n*(3/5)^n)

    now can I just take the limit from there?
     
  10. Apr 21, 2008 #9

    Dick

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    Careful, you dropped a very important '1'. Now use log(a*b)=log(a)+log(b). Then the power rule again.
     
  11. Apr 21, 2008 #10
    I came up with Limit as n tends to infinity of (ln(3)*(3/8)^n + ln(5)*(5/8)^n)
    and with those being geometric sequences with r < 1, the sequence is convergent.
     
  12. Apr 21, 2008 #11

    Dick

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    That's a long way from being correct. How did you get that?
     
  13. Apr 21, 2008 #12
    I took the natural log and I got:

    Lim 1/n * ln(3^n+5^n)

    Then applying L'hospitals:

    (3^n*ln(3)+5^n*ln(5))/(3^n+5^n)

    then I simplified from there.
     
  14. Apr 21, 2008 #13

    Dick

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    If you want to go that way then fine. Then the 'simplification' went wrong. Try dividing numerator and denominator by 5^n. What's the limit of 3^n/5^n?
     
  15. Apr 21, 2008 #14
    ok. The limit of 3^n/5^n = 0

    So after dividing through by 5^n, I'm left with ln(5). Which would be a divergent series
     
  16. Apr 21, 2008 #15

    Dick

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    It's a divergent series, but it's a convergent sequence. It converges to ln(5). But now remember you took the log of the original sequence.
     
  17. Apr 21, 2008 #16
    Oh ok. So raising ln(5) to power e gives me 5. So sequence converges to 5.
    Thanks alot for the help!
     
  18. Apr 21, 2008 #17

    Dick

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    You're welcome. I think it would be a good exercise to try and do this without l'Hopital. You don't really need it, just use the rules of logs.
     
  19. Oct 23, 2008 #18

    jlu

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    how can i proof that ([sin0.4n][/npi])2 converges and ([sin0.4n][/npi]) diverges? n is between -infinity and +infinity
     
  20. Oct 23, 2008 #19
    What is the sequence? It isn't clear in the way you've written down.
     
  21. Oct 23, 2008 #20

    Dick

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    Open a new thread for a new question. Don't tag it onto an old thread. It won't get the attention it deserves. And try to post the question more legibly on the new thread. What is ([sin0.4n][/npi])2? Do you mean (sin(theta*4n)/(n*pi))^2??
     
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