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Infinite Growth

  1. Dec 17, 2009 #1
    1. The problem statement, all variables and given/known data
    It's an extra credit assignment assigned by my math teacher, I sort of tried it but I'm not very good, and not too sure what else to do, this part isn't my strong suit.
    Here's the problem:
    "If my son would live forever and he grows 1 inch this week, 1/2 inch next week, 1/3 inch the week after that, ...etc. Then how old would my son be when he finally reaches 10 feet tall?
    (Note: He's already 8 years old and 54 inches tall.)
    (B) "If my son grows 1 inch this week, 1/4 inch next week, 1/9 inch the next week afte that, ...etc. Then what is the limit to how tall my son will grow (ie. What height will he get closer and closer to but never attain even if he lives forever?)
    2. Relevant equations

    3. The attempt at a solution

    I know it has to do with factorial (n!) and the n=1, 1/n and that side ways M thing *I forget the name*.

    I tried...(Not sure how I can write that symbol)
    Sn = /n!
    1/n = 1+1/2+1/3+1/4...+1/n. 1 3 11 50
    S(1) = 1
    S(2) = 1 + 1/2 = 3/2 2x3x-3
    S(3) = 1 + 1/2 + 1/3 = 11/6 4x3-1
    S(4) = 1 + 1/2 + 1/3 + 1/9 = 35/18

    So, I know the bottom is just n!, but I'm not sure about the top.

    Besides that I'm not really too sure what to do. Help please!! Thank you in Advance =].
    Sorry I don't have much more work to show, just not too sure where to go.
  2. jcsd
  3. Dec 17, 2009 #2


    Staff: Mentor

    Sigma is the name of this Greek letter - [itex]\Sigma[/itex]
    What's the purpose of 1, 3, 11, 50?
    I think this is what you're trying to do. Here S(n) is the sum of the first n terms, 1 + 1/2 + 1/3 + ... + 1/n.
    [tex]S(n)~=~\sum_{k = 1}^n \frac{1}{k}~=~1 + 1/2 + 1/3 + ... + 1/n[/tex]
    The common denominator for the terms in the expanded summation is n!, but I'm not sure how you can use that fact.
  4. Dec 20, 2009 #3


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    Science Advisor
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    Gold Member

    It's going to be a long time. He has 66 more inches to go. Your sum:

    Sn= 1 + 1/2 + ... + 1/n

    doesn't have a nice simple closed form But it can be shown to be greater that ln(n). How large does n need to be for ln(n) > 66?
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