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Proving {x^2} & {-n} do not converge.

  1. Sep 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove the sequences [tex]\left\{ n^{2} \right} _{n \in N}[/tex] and [tex]\left{ -n \right}_{n \in N}[/tex] do not converge.

    2. Relevant equations

    lim n^2 = a
    lim -n = b

    3. The attempt at a solution

    If n^2 converges, then for all epsilon > 0, there is an N s.t. |n^2 - a| < epsilon for all n > N.

    If -n converges, then for all epsilon > 0, there is an N s.t. |-n - b| < epsilon for all n > N.

    I assume I could use the triangle inequality possibly.
     
  2. jcsd
  3. Sep 9, 2010 #2

    jgens

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    Gold Member

    Why not just pick some concrete value for [itex]\varepsilon[/itex] (say, 1) and derive a contradiction?
     
  4. Sep 9, 2010 #3

    hunt_mat

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    Homework Helper

    If a sequence diverges then there is an M>0 thuch that when [tex]n>N[/tex] then [tex]|a_{n}|\geqslant M[/tex]. Can you do this for the sequences you have?
     
  5. Sep 9, 2010 #4

    HallsofIvy

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    Be carful. In general a sequence "diverges" as long as it does not converge. That is very different from "diverges to infinity" or "diverges to negative infinity". The sequence 1, 2, 1, 2, 1, 2, ..., diverges but does NOT satify the condition you give.
     
  6. Sep 10, 2010 #5

    hunt_mat

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    But in the context here the definition of diverges to infinity is the correct one to use.
     
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