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Homework Help: Sequences proof problem

  1. Sep 11, 2010 #1

    jgens

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    1. The problem statement, all variables and given/known data

    [tex]\lim_{n\to\infty}a_n=l \rightarrow \lim_{n\to\infty}\frac{a_1+\dots+a_n}{n}=l[/tex]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Could someone verify that this proof works? I would really appreciate it.

    Proof: Since the sequence [itex]\{a_n\}[/itex] converges to [itex]l[/itex], for any given [itex]\varepsilon>0[/itex] it's possible to find a number [itex]N>0[/itex] such that if [itex]n>N[/itex], then [itex]|a_n-l|<\varepsilon/2[/itex]. Now, because there are only finitely many numbers [itex]|a_1-l|,\dots,|a_N-l|[/itex], we can choose the greatest such number. Denote this number by [itex]M[/itex].

    Suppose that [itex]n>\max{(N,\frac{2MN}{\varepsilon})}[/itex], in which case, it clearly follows that [itex]\frac{\varepsilon}{2}>\frac{MN}{n}[/itex]. Therefore,

    [tex]\left| \frac{a_1+\dots+a_N}{n}-\frac{Nl}{n}\right|\leq\frac{|a_1-l|}{n}+\dots+\frac{|a_N-l|}{n}\leq\frac{MN}{n}<\frac{\varepsilon}{2}[/tex]

    Moreover, since [itex]n>N[/itex], we also have that

    [tex]\left| \frac{a_{N+1}+\dots+a_n}{n}-\frac{(n-N)l}{n}\right|\leq\frac{|a_{N+1}-l|}{n}+\dots+\frac{|a_n-l|}{n}<\frac{(n-N)\varepsilon}{2n}<\frac{\varepsilon}{2}[/tex]

    Combining these two results, we see that

    [tex]\left| \frac{a_1+\dots+a_n}{n}-l\right|\leq\left| \frac{a_1+\dots+a_N}{n}-\frac{Nl}{n}\right|+\left| \frac{a_{N+1}+\dots+a_n}{n}-\frac{(n-N)l}{n}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon[/tex]

    Completing the proof.
     
  2. jcsd
  3. Sep 12, 2010 #2

    jgens

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    Re: Sequences

    Bump.
     
  4. Sep 12, 2010 #3
    Re: Sequences

    Looks good to me.
     
  5. Sep 12, 2010 #4

    jgens

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    Gold Member

    Re: Sequences

    Thanks!
     
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