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Limits of infinite sums of sequences

  1. Feb 13, 2009 #1
    I understand that the limit of the sum of two sequences equals the sum of the sequences' limis: [tex]\displaysyle \lim_{n\rightarrow\infty} (a_{n} + b_{n}) = \lim_{n\rightarrow\infty}a_{n} + \lim_{n\rightarrow\infty}b_{n}[/tex]. Similar results consequenly hold for sums of three sequences, four sequences, etc. (Call this the "finite distributive property.") But what about the sum of countably many sequences? Given the sequences [tex] \{a_{i} (n)\} [/tex], is it the case that

    [tex]\displaystyle \lim_{n\rightarrow\infty} \sum_{i=1}^{\infty} a_{i}(n) = \sum_{i=1}^{\infty} \lim_{n\rightarrow\infty}a_{i}(n) ?[/tex]

    If not, are there any extra conditions on the sequences sufficient to make the equality hold? Any necessary and sufficient ones? I have an argument that if if both sides of the equality are finite, then the it holds. However, I've very likely made a mistake and would appreciate it if you guys could/would point out any errors.

    Let [tex]\displaystyle L := \Sigma_{i=1}^{\infty} \lim_{n\rightarrow\infty}a_{i}(n)[/tex]. So [tex]\displaysyle \forall \epsilon > 0 \exists S \forall s \geq S \ |L - \Sigma_{i=1}^{s}\lim_{n\rightarrow\infty}a_{i}(n)| < \frac{\epsilon}{2}.[/tex] Also let [tex]L_{n} := \Sigma_{i=1}^{\infty}a_{i}(n)[/tex]. We now therefore have, given the same [tex]\frac{\epsilon}{2}[/tex] as before, [tex]\forall T \exists t \geq T \ |L_{n} - \Sigma_{i=1}^{t} a_{i}(n) | < \frac{\epsilon}{2}[/tex]. Taking n to the limit on both sides yields [tex]\displaystyle \lim_{n\rightarrow\infty} |L_{n} - \Sigma_{i=1}^{t} a_{i}(n) | = |\lim_{n\rightarrow\infty}L_{n} - \Sigma_{i=1}^{t} \lim_{n\rightarrow\infty} a_{i}(n) | \leq \frac{\epsilon}{2}[/tex] (by the finite distributive property). Finally, define [tex]L_{0} := \lim_{n\rightarrow\infty}L_{n}, \ U := \max(S, T)[/tex]. Then for each [tex] u \geq U [/tex], we have both [tex] |L - \Sigma_{i=1}^{u} \lim_{n\rightarrow\infty} a_{i}(n) | < \frac{\epsilon}{2}[/tex] and [tex]|L_{0} - \Sigma_{i=1}^{u} \lim_{n\rightarrow\infty} a_{i}(n) | \leq \frac{\epsilon}{2}[/tex]. By the triangle inequality, then, [tex] |L - L_{0}| < \epsilon[/tex]. Since epsilon was arbitrary, [tex]L = L_{0}[/tex].

    What about the cases where the sides aren't finite?
     
    Last edited: Feb 13, 2009
  2. jcsd
  3. Feb 13, 2009 #2
    I don't know the answer off the top of my head, but I do want to point this out.

    What you have is essentially TWO limits. The infinite sum cannot be directly evaluated and so it is actually a limit in disguise. So your question boils down to when can you interchange two limit operations?

    I'm not a good analysis student, though, and I don't know the exact conditions.
     
  4. Feb 13, 2009 #3

    lurflurf

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    necessary and sufficient conditions are hard to come by
    uniform convergence is a useful though limited sufficient condition
     
  5. Feb 13, 2009 #4
    O.K., here's an easy counterexample to the general claim:

    [tex]\displaystyle \sum_{i=1}^{\infty} \lim_{n\rightarrow\infty} \frac{1}{2^{n}} = 0, \ \lim_{n\rightarrow\infty} \sum_{i=1}^{\infty} \frac{1}{2^{n}} = \infty. [/tex]

    Limits usually behave so naturally in terms of commuting with other operators that I'm still somewhat surprised they don't commute here.
     
  6. Feb 13, 2009 #5

    lurflurf

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    Limits usually behave so naturally in terms of not commuting with other operators that I'm still somewhat unsurprised they don't commute here. Often uniform situations are used that gives commutation. It is easy to produce counter examples by using nonuniform limits.
     
  7. May 19, 2011 #6
    Hey mag487,

    Did you get to verify if your argument was correct? I went through your proof and couldn't find any loophole; however I am no expert in analysis.
     
  8. May 20, 2011 #7
    Since the 1/2^n doesn't have an i in it you can take it out of the summation in both cases so you get 0*infinity in both cases. If you adopt a 0* infinity =0 convention then both sides are equal to zero.
     
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