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Proof of the preliminary test (Divergence test for series)

  1. Jun 19, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]Proove \, that \, if \, \lim_{n \to \infty} a_{n} \neq 0 \, , \, \sum^\infty a_{n} \, diverges[/tex]


    2. Relevant equations
    [tex] S_{n} - S_{n-1} = a_{n}[/tex]
    [tex] \lim_{n \to \infty} S_{n} = S[/tex]

    3. The attempt at a solution
    [tex] S_{n} - S_{n-1} = a_{n}[/tex]
    [tex] \lim_{n \to \infty} S_{n} - S_{n-1} = \lim_{n \to \infty} a_{n}[/tex]
    Since [itex] a_{n} \neq 0 [/itex], we can divide by [itex]a_{n}[/itex]

    [tex] \lim_{n \to \infty} \frac {S_{n} - S_{n-1}} {a_{n}} = 1[/tex]
    We can rewrite [itex] S_{n} [/itex] and [itex] S_{n-1} [/itex] as [itex] \sum^\infty a_{n} [/itex] and [itex] \sum^\infty a_{n-1} [/itex]

    So, [tex] \lim_{n \to \infty} \frac {\sum^\infty a_{n} - \sum^\infty a_{n-1}} {a_{n}} = 1[/tex]
    Rearranging, [tex] \lim_{n \to \infty} \frac{\sum^\infty a_{n}} {a_{n}} - \sum^\infty \frac{a_{n-1}} {a_{n}} = 1[/tex]
    The term inside the second series is 1, so
    [tex] \lim_{n \to \infty} \frac{\sum^\infty a_{n}} {a_{n}} = 1 + \sum^\infty 1 [/tex]

    Thus, [tex] S_{n} = \lim_{n \to \infty} \sum^\infty a_{n} = a_{n} (1 + \sum^\infty 1) [/tex]
    Since [itex] \sum^\infty 1[/itex] diverges, we can say that [itex] S_{n} [/itex] diverges and therefore, if [itex]\lim_{n \to \infty} a_{n} \neq 0 [/itex] , [itex] \sum^\infty a_{n} [/itex] diverges.

    Is this proof complete?
     
    Last edited: Jun 19, 2013
  2. jcsd
  3. Jun 19, 2013 #2

    LCKurtz

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    That isn't a correct statement of the theorem in the first place. When you say ##\lim_{n \to \infty} a_{n} \neq 0## you are implying ##a_n## has a limit which isn't ##0##. That is not the correct denial of the statement that ##a_n\rightarrow 0##. It may not converge to anything.

    You can't take a limit since it isn't given ##a_n## has one. And nothing gives you ##a_n\ne 0##.

    I would suggest you try the showing the contrapositive: If ##\sum a_n## converges then ##a_n\rightarrow 0##.
     
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