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Proving null sequences

  1. Apr 4, 2012 #1
    I know in general how to prove if sequence is null or not . But here is my confusion
    - method in the text book I am reading - asks that to prove that any sequence is null we must show that for each ε>0, there is an integer N such that modulus of the given sequence is < ε, for all n>N

    now I also understand this but my dilema is
    taking modulu say for example given sequence is
    {(1+(-1)^n)/(n+(-1)^n )}
    taking modulus gives:
    |(1+(-1)^n)/(n+(-1)^n )| is modulus 2/(n+1) ???

    just not sure
  2. jcsd
  3. Apr 4, 2012 #2

    [itex]\left|\frac{1+(-1)^n}{n+(-1)^n}\right|=\left\{\begin{array}{cc}0\,&\,if\,\,n\,\,is\,\,odd\\\frac{2}{n+1}\,&\,if\,\,n\,\,is\,\,even\end{array}\right.[/itex] , and of course the seq. begin with [itex]n=2[/itex].


    Ps. Of course, the absolute value has no relevance here...did you notice this?
  4. Apr 4, 2012 #3
    Yes the question states n=2,3... So I am correct and must ignore negative values of n
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