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Is my proof for Absolute convergence => Convergence right?

  1. Mar 29, 2010 #1
    Is my proof for "Absolute convergence => Convergence" right?

    I had to prove it for my calc test today; my proof looks different than the one my prof posted as the answer; it's also way shorter. He used the one from the book but I didn't memorize it so I used this. Is it right?

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

    Prove that Absolute Convergence implies convergence


    3. The attempt at a solution

    Notation:

    1. I'm using the notation an means a series. [As in: an = a(1)+a(2)+a(3)...+a(n) and n goes to infinity]

    2. |an| means the absolute value of |an|.

    ***

    I. |(-1)^n an| = an.

    II. Suppose an converges. i.e: an = C

    III. an converges implies that a(n+1) < a(n)

    IV. (III.) implies (-1)^n an converges.

    QED.
     
    Last edited: Mar 29, 2010
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  3. Mar 29, 2010 #2

    jbunniii

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    Re: Is my proof for "Absolute convergence => Convergence" right?

    There are a lot of problems with this.

    (1) Your notation is inconsistent. In item (I), do you mean an or a(n)?

    (2) (II) is false as stated. You are missing a "[itex]\lim_{n\rightarrow\infty}[/itex]".

    (3) (III) is false. The terms of a series do not have to be monotonically decreasing in order for the series to converge. (They do have to converge to zero, though.)

    (4) In (IV) you seem to be assuming that the series is alternating. This is one special case. But absolute convergence implies convergence for ANY series, not just alternating series. Also once again I assume you mean a(n) instead of an. By the way, a more standard notation would be [itex]s(n) = a(1) + \ldots + a(n)[/itex].

    I recommend carefully reading the proof posted by your prof.
     
  4. Mar 29, 2010 #3

    Fredrik

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    Re: Is my proof for "Absolute convergence => Convergence" right?

    In (I), you're using an to mean two different things, and you have already confused me enough to not understand what you mean. In (II), you're making an assumption that appears to be irrelevant to the problem. Did you leave out some details? (III) is wrong. (IV) I'm not sure what you mean, but this looks wrong too.
     
  5. Mar 29, 2010 #4
    Re: Is my proof for "Absolute convergence => Convergence" right?

    Yeah, as I'm reading this I see I should have clarified the following, which I assumed needn't be clarified:

    I mean an. As in, the absolute value of an alternating series is the same as the value of that series if it were non-alternating. (because an alternating series is essentially (-1)^n an.)

    Why? -- an is the sum of the series in this notation as n goes to infinity. That's what my definition says.

    This was an actual mistake. I meant to write |a(n+1)| < |a(n)|

    I also did not explain that if III is not satisfied by a series that goes negative (for example, 1/1, -1/2, 0, 0, 1/3, 0, -1/4; or a series involving something like sin(n), which goes negative and positive but is not an alternating series, then the proof still holds because such a series can always be expressed as smaller than a series that does satisfy III. For example, the one involving sin(n) can be expressed as a smaller convergent series because |sin(n)| ≤ 1)

    I guess I should have included all those observations in my proof?

    IV simply uses the definition of a convergent alternating series (that if a(n + 1) < a(n), it converges).

    Damn, was it really not easy to follow? it was so clear in my mind lol -- Other than my broad assumptions, did I make any actual mistakes? Given my explanation, is the proof valid?
     
    Last edited: Mar 29, 2010
  6. Mar 29, 2010 #5
    Re: Is my proof for "Absolute convergence => Convergence" right?

    O, and I used proper sigma notation on my test. I just used weird notation here on PF to make things easier for myself on the computer (I'm not very comfortable with laTex).

    So try and disregard the weird notation.
     
  7. Mar 29, 2010 #6

    Fredrik

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    Re: Is my proof for "Absolute convergence => Convergence" right?

    That doesn't actually make sense, because it would mean [tex](-1)^n\sum_{k=1}^\infty a_k[/itex], but I understand that you mean [tex]\sum_{k=1}^\infty(-1)^k a_k[/tex].

    This still isn't true. For example, |a571| doesn't have to be smaller than |a570|.

    Tip: Click the quote button to see how I typed the above.
     
  8. Mar 29, 2010 #7
    Re: Is my proof for "Absolute convergence => Convergence" right?

    lol yes.

    That's what I meant. If the series doesn't have that property, like one that involves sin(n), such a series can still be expressed as one that is smaller than or equal to one that does have that property, or one that is convergent; and so it itself must be convergent.

    For example, 1/2, 1/2, -1/3, -1/3, ... can be expressed as twice the alternating series 1/2, -1/3, ...

    Or the series 1/2, 1/2, -1/3, -1/4, ... can be expressed as the series 1/2, -1/3... + the series 1/2, -1/4...

    And the series sin(n)/n^p ≤ 1/n^p .

    etc.

    That's probably the missing step I should have added to seal the proof... unless you still think it doesn't hold?

    It's what I had on my mind when I wrote it, but we only had 10 minutes per question, and it didn't occur to me to clarify that. It's the mindreader syndrome, lol; you think that because it's what you have in mind, the person reading it will see it that way.
     
    Last edited: Mar 29, 2010
  9. Mar 29, 2010 #8

    jbunniii

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    Re: Is my proof for "Absolute convergence => Convergence" right?

    So you are asserting that

    [tex]\left|\sum_{n=1}^{\infty}(-1)^n a_n\right| = \sum_{n=1}^{\infty} a_n[/tex]

    or what? Because that is certainly not true.


    Well, from what you wrote I assumed you meant the n'th partial sum of the series. If an is the limit of the partial sums, then it does not depend on n, so your notation is extremely bad.

    I assume you mean that there exists a monotonically decreasing subsequence

    [tex]a_{n_k}[/tex]

    Whether or not this is true (can you prove it?), you can't conclude that

    [tex]\sum_{n}a_n[/tex]

    converges simply because

    [tex]\sum_{k}a_{n_k}[/tex]

    converges.

    By the way, neither

    [tex]\sum_n |\sin(n)|[/tex]

    nor

    [tex]\sum_n \sin(n)[/tex]

    converges. The terms don't even converge to 0.

    No, I don't think it's valid at all.
     
  10. Mar 29, 2010 #9
    Re: Is my proof for "Absolute convergence => Convergence" right?

    I'm talking about absolute convergence, which means:

    [tex]\sum_{n=1}^{\infty}\|\(-1)^n a_n\| = \sum_{n=1}^{\infty} a_n[/tex]

    Which is true. Isn't it?

    (I have no idea how to write it properly on laTex and I'm tired and on a train. it's supposed to be ∑|(-1)^1 an| )



    I think you misunderstand what I wrote.

    See my previous comment:

    I know sin(n) doesn't converge. But a series containing sin(n) may converge; that's all I was talking about. I gave a series containing sin(n) as an example of a non-alternating series that sometimes goes negative.
     
  11. Mar 29, 2010 #10

    Fredrik

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    Re: Is my proof for "Absolute convergence => Convergence" right?

    There are few too many backslashes in your LaTeX. The code for the left-hand side is \sum_{n=1}^\infty|(-1)^n a_n|, but absolute convergence means

    [tex]\sum_{n=1}^\infty |a_n|<\infty[/tex]

    not

    [tex]\sum_{n=1}^{\infty}|(-1)^n a_n|=\sum_{n=1}^\infty a_n[/tex]

    You shouldn't be rearranging terms if you don't know exactly what you're doing. For example 1-1+1-1+1-1+1-1+...=0, right?, but if you rearrange the terms, 1+1-1+1+1-1+1+1-1+...=1+1+1+1+...=∞.

    I think you should probably start over, and be more careful with the definitions and the notation this time.
     
  12. Mar 29, 2010 #11
    Re: Is my proof for "Absolute convergence => Convergence" right?

    lol yeah I think I should rewrite the whole thing from the start carefully. I might do that tomorrow. I just arrived home, I was really tired and after 5 hours sitting on a train and a day without sleep I don't really have the energy to be rigorous (or even legible apparently :rofl:)

    Anyway, thanks!
     
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