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How to analytically prove a sum goes to zero

  1. Dec 11, 2011 #1
    How would you analytically prove that:

    [itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\frac{1}{m}[/itex][itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex])= 0 ?


    The way I did it, I just proved that the greatest lower bound of [itex]\frac{1}{m}[/itex] is 0, and so since the function is monotonically decreasing I proved that the limit of [itex]\frac{1}{m}[/itex] is 0, so anything multiplied by 0 (like the limit of [itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex]) must also be 0.

    My professor agreed with me.

    But now that I look at it again, obviously it can't be right, because the limit of [itex]\frac{1}{m}[/itex][itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex] as m goes to infinity is infinity, and 0 times infinity is indeterminate...

    But my professor seems really smart and paid great attention to my work...is he wrong?














    (My exact work was as follows):
    [itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\frac{1}{m}[/itex][itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex]) = [itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\frac{1}{m}[/itex])*[itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex])
    Let b be the greatest lower bound of [itex]\frac{1}{m}[/itex]. If b=0, then:
    [itex]\frac{1}{m}[/itex]>0 for all m
    Since b+ε cannot be a greatest lower bound, then:
    0+ε>[itex]\frac{1}{N}[/itex] for some N
    if m≥N, then f(N)≥[itex]\frac{1}{m}[/itex], since the function is monotonically decreasing. So:
    0+ε>f(N)>[itex]\frac{1}{m}[/itex]>0
    Therefore:
    ε>[itex]\frac{1}{m}[/itex]>0, where ε>0 can be made arbitrarily close to 0.
    Therefore, by the squeeze theorem, the limit of [itex]\frac{1}{m}[/itex] as m→∞ is 0.
    Therefore,
    [itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\frac{1}{m}[/itex])*[itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex])=0
     
  2. jcsd
  3. Dec 11, 2011 #2

    mathman

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    Science Advisor
    Gold Member

    I haven't analyzed your proof. However, the sum ~ ln(m), so the expression is ~ ln(m)/m -> 0
     
  4. Dec 11, 2011 #3
    Yes I know that but ln(m)→∞ as m→∞...so either way you have an indeterminate form...
     
  5. Dec 11, 2011 #4
    Is my question stupid or does no one know the answer????
     
  6. Dec 11, 2011 #5

    Mute

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    Homework Helper

    "Indeterminate form" does not mean the expression does not have a limit, it just means that the limit, if it exists, cannot be found by directly plugging in "m = infinity" into your expression.

    log(m)/m is indeterminate, but the value of the limit can be found using L'Hopital's rule.

    Your original proof was incorrect, it seems, because you didn't actually evaluate the indeterminate form, you just figured out one part of the expression had the limit of zero and wrote that as the answer. In this case, zero happened to be the answer, but for different reasons.

    To really prove the limit, you could use the integral test to put an upper bound on the sum, and then show the limit of the upper bound is zero (which will require you to use L'Hopital's rule). This is essentially what mathman told you to do.
     
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