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Mean of an Infinite Sequence

  1. Aug 27, 2010 #1

    Char. Limit

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    1. The problem statement, all variables and given/known data
    So, I was thinking about the mean of a sequence the other day, and you know how the mean of a sequence is usually written (I believe) like this?

    [tex]\frac{\sum_{i=1}^n a_i}{n}[/tex]

    So I was considering the mean of an infinite sequence, and I came up with an example. Now I want your help to see if I'm evaluating it right.


    2. Relevant equations
    [tex]\frac{d}{dx}[f_1(x) + f_2(x) + ... + f_n(x)] = \frac{d}{dx}f_1(x) + \frac{d}{dx}f_2(x) + ... + \frac{d}{dx}f_n(x)[/tex]

    In other words, the derivative of a sum is the sum of the derivatives.

    3. The attempt at a solution

    So the sequence that I picked is:

    [tex]a_i = \frac{i-1}{i}[/tex]

    Thus, I believe the mean of every number in this sequence would be...

    [tex]lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n \frac{i-1}{i}}{n} = lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n 1 - \frac{1}{i}}{n}[/tex]

    Because this limit, if directly evaluated, would yield infinity/infinity, I'll try using L'Hopital's rule and the rule that says the derivative of a sum is the sum of the derivatives...

    [tex]lim_{n\rightarrow\infty} \frac{\frac{d}{dn}\sum_{i=1}^n 1 - \frac{1}{i}}{\frac{d}{dn} n} = lim_{n\rightarrow\infty} \frac{\sum_{i=1}^n \frac{d}{di}\left(1-\frac{1}{i}\right)}{\frac{d}{dn} n}[/tex]

    So, these derivatives can be evaluated, and so I will, yielding...

    [tex]lim_{n\rightarrow\infty}\frac{\sum_{i=1}^n \frac{1}{i^2}}{1}[/tex]

    So, I just need to find what the infinite sum from 1 to infinity of 1/i^2 is.

    Using Wolfram-Alpha, I generate the number...

    [tex]\frac{\pi^2}{6} = 1.64493...[/tex]

    So, how did I generate, from a sum of a sequence of numbers from 0 (n-1/n for n=1) to 1 (the limit as n approaches infinity), a mean greater than any of the numbers? Is there a mistake here?
     
  2. jcsd
  3. Aug 27, 2010 #2

    jgens

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    I'm fairly certain that this step isn't justified.
     
  4. Aug 27, 2010 #3

    Char. Limit

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    Oh, I see. So I'm guessing that I have to instead first evaluate the partial sum, then differentiate it? Or do I even have to do the latter? Perhaps I can just divide the partial sum by n and then take the limit.

    Just wondering, how exactly can you evaluate the partial sum of 1/i?
     
  5. Aug 27, 2010 #4

    diazona

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    Yeah, you have to evaluate the partial sum before taking any derivatives, because inside the sum, n isn't even a defined variable.

    The way I would do it doesn't involve any derivatives at all, it basically does come down to dividing the partial sum by n and then taking the limit. The first term is easy, of course, but for the second term you wind up with
    [tex]\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{i}[/tex]
    The result of that sum (without the 1/n in front) is called the n'th harmonic number. Since you're taking the limit as n goes to infinity, you can get an approximation for the value of the sum (at large n) by converting it to an integral. Or you can look up the properties of harmonic numbers in your favorite math reference, including approximations for the sequence at large values of n.
     
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