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Proving the Divergence of a Sequence

  1. Mar 29, 2004 #1
    Hey there, one of our h/work questions is to prove that a certain sequence is divergent, where xn=(-1)^n for every natural n. I started off by assuming that it was infact convergent so wrote that mod(xn/l)<e where e is any real number greater than zero, and this holds for any n>no. But from the one example we did in class, it seems you can choose your e to make things a little easier for yourself and then choose k and k+1 both greater than no to allow you to obtain the contradiction that you need. I mean I started to try things like mod(xn/l)<1 and then choose mod(xk-l)<1 and (xk+2-1)<1. Then I had 2=mod((k+2)-k) which was equal to mod(xk+2 +xk) but this clearly doesn't seem to get me anywhere (well not from where I'm sitting). I've been trying stuff like this but can't seem to get anything to work. Could someone please put me in the right direction? Thanks!

    Btw, when I've written stuff like xk+2 I just mean xsubscript k+2 :smile:
  2. jcsd
  3. Mar 29, 2004 #2

    matt grime

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    it has a convergent subsequence - for n even the subsequence tends to 1. It also has the odd terms converging to -1. thus the sequence cannot converge or any two subsequences must converge and converge to the same thing. No need to use epsilon, but we can do it that way if we must, it's just messy, the start there is to suppose ti converges to x, if x is positive do something with e= 1/2 and the odd terms, if x is -ve do it with e=1/2 and the even terms.
  4. Mar 29, 2004 #3
    Pfft why do they give us a messy one to prove? We're only analysis babies!

    So I've said mod(xn-l)<1/2 for all n>no, so l is what it's converging to (just what I'm used too).
    Now we take mod(xk-l)<1/2 and mod(xk+1)<1/2

    Taking the situation first where k is an even power so the limit is negative we have

    mod(1+l)<1/2 and mod(-1+l)<1/2

    Then have 1=mod((1+l)-l)<mod((1+l)+(-l+1))<mod(1+l)+mod(-l+1)<1/2 +1/2<1 which obv doesn't make sense, but I think this is v.wrong. Ooops.
  5. Mar 29, 2004 #4

    matt grime

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    Remember you order of qunatifiers.

    to show x(n) converges means

    there is an x

    such that

    for all e>0

    there is an N(e)

    such that

    for all n>N(e)

    |x(n) - x| <e

    To show it does not converge

    for all x

    there is an e

    such that

    for all N(e)

    there is an n>N(e)

    with |x(n)-x| >e

    So to show it does not converge (without using our brains) we say, suppose x is arbitary.

    case 1 x positive, let e =1/2, then for any N, take the next odd number greater than N, say it is n, then |x_n-x| >1/2 cos x_n is -1 and every postive number is more than 1/2 away from -1

    case 2 if x is negative, let e=1/2, then for any N take the next even number after N, call it n, then |x_n-x|>1/2 cos every negative number is mroe than 1/2 away from 1.
  6. Mar 29, 2004 #5


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    If you want to be formal:

    Assume (by contradiction) that the series converges to some value [tex]s[/tex]. Then given any [tex]\epsilon > 0[/tex] there must be some [tex]N[/tex] such that [tex]n>N \rightarrow |s-\sum_{i=1}{n}(-1}^i|<e[/tex]. Now if [tex]\epsilon<0.5[/tex] consider that [tex]|\sum_{i=1}{N+1}(-1)^i-\sum_{i=1}{N+1}(-1)^i|=1[/tex] and that [tex]\sum_{i=1}{N+1}(-1)^i = s + e_1[/tex] and [tex]\sum_{i=1}{N+1}(-1)^i=s-e_2[/tex] where [tex]|e_1|,|e_2| < \epsilon[/tex] so by the triangle inequality [tex]|e_1+e_2|<2 \epsilon < 1[/tex] but by substitution we also have [tex]|\sum_{i=1}{N+1}(-1)^i-\sum_{i=1}{N+1}(-1)^i|=|s + e_1 - (s - e_2)| = |e_1 + e_2| = 1[/tex] this is a contradiction therefore the inital assumption must be incorrect.

    If you understand this, you should have no trouble generalizing it to include all series [tex]\sum a_i[/tex] where [tex]\lim_{i \rightarrow \infty} a_i \neq 0[/tex]
  7. Mar 29, 2004 #6

    matt grime

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    but the question wasn't on series.
  8. Mar 29, 2004 #7
    Okies, thanks for that I'm pleasantly surprised by how much simpler it was than I first thought!

    Just one more ickle quiestion. If we have xn tending to l as n tends to infinity, how do we prove that modxn tends to mod(l) as n tends to infinity.

    So we have mod(xn-l)<e for all n>N

    Then I put mod(xn-l)>mod(xn)-mod(l) although that was a tad pointless I think.... Then I put mod(xn-l)>=mod(mod(xn)-mod(l)) but I'm not sure if that's valid or not....
  9. Mar 29, 2004 #8

    matt grime

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    Use the folllowing inequality:

    ||x|-|y|| < |x-y|
  10. Mar 29, 2004 #9
    Thanks, I didn't know that existed. I'd only ever seen the triangle one before, but never that one. Is there any way of proving it? Sorry, I'm being a pain. Thanks again! :wink:
  11. Mar 29, 2004 #10

    matt grime

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    It follows from the triangle inequality:

    |x| = |x-y+y| < |x-y| + |y|

    so |x|-|y| < |x-y|

    and by symmetry


    so -|x-y| <|x|-|y| < |x-y|

    ie ||x|-|y|| < |x-y|
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