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Convergence of this sequence .

  1. Jun 10, 2013 #1
    1. The problem statement, all variables and given/known data
    find the limit n[itex]\rightarrow[/itex]∞ of 10n/ n!


    2. Relevant equations
    L hospital rule


    3. The attempt at a solution
    took log and separated the num and denom as:
    n ln10-ln(n!)
    n ln10-n ln(n)+n
    1/n ( ln10 - ln(n)+1)
    now i applied l hospital rule
    then i got lim n[itex]\rightarrow[/itex]∞ as 0.So the actual answer is 1. (e0)
    I just want to know if the approach is correct.
     
  2. jcsd
  3. Jun 10, 2013 #2
    I'm afraid I don't see how you transition between these two steps.
    Also, intuitively the factorial function should dominate and the limit should be zero.
     
  4. Jun 10, 2013 #3
    my bad ,in my desperate attempt to get a ratio, I did that stupid thing. How do we solve it beyond intuition?
     
  5. Jun 10, 2013 #4
    From n ln10-n ln(n)+n, you can combine all the terms into a single expression. Taking the limit, you will find that the expression tends to minus infinity. The limit of the original expression is then 0.
     
  6. Jun 10, 2013 #5
    n*ln(10e/n) this is what i am getting , and it is not -infinity .
     
  7. Jun 10, 2013 #6

    Dick

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    Yes, the limit of that is -infinity. n goes to +infinity. What does ln(10e/n) do?
     
  8. Jun 10, 2013 #7
    I took n out , then
    n( ln10-lnn +1 )
    n(ln10-lnn+lne)
    n ln(10e/n)
    i don't know what to do after this.
     
  9. Jun 10, 2013 #8
    Are you saying that ln0 is taken as -inf, but there is also n before that right which becomes +inf.
     
  10. Jun 10, 2013 #9

    HallsofIvy

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    The ratio test gives: [itex]a_{n+1}/a_n= (10^{n+1}/(n+ 1)!)(n!/10^n)= (10^{n+1})/10^n)(n!/(n+1)!)= 10/(n+1)[/itex] goes to 0 as n goes to 0. Strictly speaking, the ratio test shows that [tex]\sum a_n[/tex] converges but if that is so, then, of course, the sequence [itex]\{a_n\}[/itex] converges to 0.

    (The converse is not necessarily true. For example, [itex]\{1/n\}[/itex] clearly converges to 0 but the sum [tex]\sum 1/n[/tex] does NOT converge so the ratio test does not work.)
     
    Last edited by a moderator: Jun 11, 2013
  11. Jun 10, 2013 #10

    Office_Shredder

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    n*ln(10e/n) if n is really big is a really big positive number multiplied by a really big negative number. What is that going to be?
     
  12. Jun 12, 2013 #11
    Now i get it , it is going to really big negative number, so -inf.
    therefor the answer for the original question is 0.
    thankyou everyone.
     
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