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Infinite sequence

  1. Jul 29, 2006 #1
    I want to find the smallest value of N (n > N) such that the following is true:

    abs((2n)^(1/n) - 1) < 0.01

    So basically, I'm looking for the term of the sequence (2n)^(1/n) which is less than 0.01 from the limit of the sequence... by trial and error, I found it to be 734.

    But I'm looking ofr a theoretical solution. Any help as to what I'm missing here?

  2. jcsd
  3. Jul 29, 2006 #2


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    Well, the sequence approaches the limit from above, so you can just write (2n)1/n<1.01. From here, I would take the log of both sides. You probably won't be able to solve for the minimum value of N exactly without a calculator, but you should be able to use some estimates to get an upper bound that isn't too bad.
  4. Jul 29, 2006 #3
    So you're looking for the value n so :

    [tex] |(2n)^{1/n}-1|<0.01 [/tex] then i must be satisfied that.

    [tex] (2n)^{1/n}<1.01 [/tex] taking log at both sides you have:

    [tex] \frac{log2}{n^2 }+\frac{logn}{n^2 }<log(1.01)=0.00995033 [/tex]

    for big n n>100 we can ignore the first term and have:

    [tex] log(n) < n^2(0.00995033) [/tex] trying several n we have that n=60
  5. Jul 29, 2006 #4
    Hmmm... (2*60)^(1/60) - 1 is approx 0.08

    I think 733 is the answer, but I still think there must be another way besides trial and error / using the calculator
  6. Jul 29, 2006 #5


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    Actually, the easiest way to do it by hand is probably to raise both sides to the nth power and then expand the RHS:


    [tex]2n<(1.01)^n=(1+\frac{1}{100})^n=1+n \cdot \frac{1}{100}+\frac{n(n-1)}{2} \cdot \frac{1}{100^2}+...[/tex]

    Now, if you replace the RHS by something that is smaller than it, and find an n such that the LHS is smaller than this new RHS, you know the LHS will also be smaller than the original RHS at this n. One way to do this is to truncate the series after the third term:

    [tex]2n<1+n \cdot \frac{1}{100}+\frac{n(n-1)}{2}\cdot \frac{1}{100^2}\leq(1.01)^n[/tex]

    Now solve for the smallest n that satisfies this first inequality and you'll get an upper bound on the smallest n for the original inequality. If you wanted to get better accuracy you could add more terms, but this would become harder to solve.
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