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Homework Help: Rigorous proof of limits of sequences (2)

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Definition: Let an be a sequence of real numbers. Then an->a iff
    for all ε>0, there exists an integer N such that n≥N => |an - a|<ε.

    [for all of the following, "lim" means the limit as n->∞]
    Theorem: Suppose lim an =a and lim bn =b. Then lim (an + bn) = a + b.

    Proof:
    Given ε>0, find N1 s.t. |an -a|< ε/2 for all n≥N1 and find N2 s.t. |bn -b|< ε/2 for all n≥N2.
    Let N=max{N1,N2}. Then if n≥N,
    |(an +bn) - (a+b)| = |(an -a) + (bn -b)| ≤ |an -a| + |bn -b| ≤ ε/2 + ε/2 = ε.
    =========================

    I am very confused about this proof and I'm never able to completely understand it since first year calculus.
    In the definition, we have ε, but in this proof, WHY is it valid to take ε/2 instead of ε? What is the core of the reason that allows us to do this? I don't follow the logical flow of the argument. To me, taking it to be ε/2 instead of ε is like cheating...

    2. Relevant equations
    N/A

    3. The attempt at a solution
    N/A

    I hope someone can explain this in more detail.
    Thank you! :)
     
    Last edited: Jan 8, 2010
  2. jcsd
  3. Jan 8, 2010 #2
    the reason they use one half epsilon is to not confuse people not the other way around. the point is that from the original definition epsilon was arbitrary small. so lets say you are doing the second proof and instead of using epsilon start using delta. no give any delta you want to prove
    Given &>0, find N1 s.t. |an -a|< ε for all n≥N1 and find N2 s.t. |bn -b|< ε' for all n≥N2. Let N=max{N1,N2}. Then if n≥N,
    |(an +bn) - (a+b)| = |(an -a) + (bn -b)| ≤ |an -a| + |bn -b| ≤ ε + ε' < &.
    so given any & you can choose an epsilon and epsilon prime such that the inequality holds. in the give proof above you picked a special epsilon and epsilon prime such that their sum is the given delta.
     
  4. Jan 8, 2010 #3

    vela

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    The first [tex]\epsilon[/tex] is for the combined sequence. We assume it's given.

    Because the sequence [tex]\{a_n\}[/tex] converges to [tex]a[/tex], we know that for every [tex]\epsilon_a>0[/tex], there exists an [tex]N_1[/tex] such that...blah blah blah. In particular, you can find an [tex]N_1[/tex] for when [tex]\epsilon_a = \epsilon/2[/tex].

    Similarly, because the sequence [tex]\{b_n\}[/tex] converges, we know that for every [tex]\epsilon_b>0[/tex], there exists an [tex]N_2[/tex] such that...blah blah blah. In particular, you can find an [tex]N_2[/tex] for when [tex]\epsilon_b = \epsilon/2[/tex].

    You choose these values for [tex]\epsilon_a[/tex] and [tex]\epsilon_b[/tex] simply because they make the rest of the proof work.
     
  5. Jan 8, 2010 #4
    Thanks! I think this is exactly the explanation that I'm looking for.

    But I would like to double check my understanding.

    1) So an->a
    => for ALL εa>0, there exists an integer N such that n≥N => |an - a| < εa. (by definition)
    Since the statement is true for ALL εa>0, IN PARTICULAR the statement must be true for εa = ε/2 (since ε/2 is a particular GIVEN postiive number), so we know that "there exists (and we can find) an integer N such that n≥N => |an - a| < ε/2". Am I right??

    2) Instead of ε/2 and ε/2, we can as well take ε/4 and 3ε/4 instead, right?
     
  6. Jan 8, 2010 #5

    vela

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    Yes and yes.
     
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