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Limit of a Series

  1. May 31, 2015 #1
    What should be the limit of the following series (if any ...)

    [tex]\frac{{1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}}}{n}[/tex]
     
  2. jcsd
  3. May 31, 2015 #2

    wabbit

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    Technically, this is not a series but a sequence.

    That aside, one way to get the answer is to use, for ##k\geq 2, \frac{1}{k}=\int_{k-1}^k\frac{dx}{k}\leq \int_{k-1}^k \frac{dx}{x} ## .
     
  4. May 31, 2015 #3
    I have just found here that, given a sequence [tex]{\left\{ {{a_n}} \right\}_{n = 1}}^\infty [/tex], [tex]\mathop {\lim }\limits_{n \to \infty } {a_n} = L \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n}} \right) = L[/tex]

    Instant answer ¬¬'
     
  5. May 31, 2015 #4

    wabbit

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    Ah yes, even simpler you're right : )
     
  6. Jun 5, 2015 #5

    disregardthat

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    I don't see how that follows actually. Use that for every ##\epsilon > 0##, there is an ##N## such that ##|a_n-L| < \epsilon##. Then for all ##n## larger than ##N##,
    ##|\frac{a_1 + ... + a_n}{n} - L| = |\frac{a_1 + ... + a_N}{n} + \frac{(a_{N+1}-L) + ...+ (a_n-L)}{n}| \leq |\frac{a_1 + ... + a_N}{n}| + |\frac{(a_{N+1}-L)}{n}| + ...+ |\frac{(a_n-L)}{n}| < |\frac{a_1 + ... + a_N}{n}| + \frac{\epsilon}{n} + ... + \frac{\epsilon}{n} = |\frac{a_1 + ... + a_N}{n}| + \epsilon.##

    As ##n \to \infty##, the expression ##|\frac{a_1 + ... + a_N}{n}|## will converge to 0. In particular, there is some N' such that for all n larger than N', we have ## |\frac{a_1 + ... + a_N}{n}| < \epsilon##. Thus we find that choosing N' large enough (bigger than N), we will have for any chosen ##\epsilon > 0## and for all n larger than N', that ##|\frac{a_1 + ... + a_n}{n} - L| < 2\epsilon## proving that ## \lim_{n \to \infty} |\frac{a_1 + ... + a_n}{n} - L| = 0##.

    Note: edited
     
    Last edited: Jun 5, 2015
  7. Jun 6, 2015 #6
    I know this, it's called Cesaro's theorem.
    For the proof you can write:


    ## |(\frac{1}{n}\sum_{k=1}^n a_k) - L | \le \frac{1}{n}\sum_{k=1}^N |a_k - L| + \frac{1}{n}\sum_{k=N+1}^n |a_k - L| ##

    The first term tends to 0 as n tends to infinity.
    The second term is bounded by ##\frac{n-N}{n} \varepsilon ## for N big enough
     
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