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Homework Help: Advanced Calculus Proof-limit of a sequence.

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that the sequence {a_n} converges to A. Define the sequence {b_n} by:

    b_n = (a_n +a_(n+1) )/2

    Does the sequence {b_n} converge? If so, specify the limit and prove your conclusion. Otherwise, give an example when this is not true.

    2. Relevant equations

    3. The attempt at a solution

    Me and some class mates tried to use the sandwich theorem, but are not sure if it applies here:

    Suppose that {a_n}, {k_n}, and {c_n} are three sequences. NOTE: k_n = a_(n+1). Suppose there exists an element n from N, such that


    Suppose that a_n and c_n converge to A, then, by the limit definition, there exists an n_2 such that |a_n-A|<ε

    the same thing for c_n, there exists an n_3 such that |c_n-A|<ε

    n* max {n_1, n_2, n_3}

    A-ε≤a_(n )≤A+ε
    A-ε≤c_(n )≤A+ε
    then A-ε≤a_(n )≤k_n≤c_n≤A+ε so that means that K_n converges to A, but k_n is actually
    a_(n+1), so then a_(n+1) converges to A.

    Now, since b_n =( a_n +a_(n+1))/2

    take the limit of b_n and by limit preperties you find that the limit of b_n is A.

    I dont know if this is the right solution for this problem, we also thought about representing the a_n as a sequence that converges to A, such as a_n =A, but Im not sure since it seems we are asked to prove this generally.

    Any hints would be highly appreciated,

    thank you very much,

  2. jcsd
  3. Sep 28, 2008 #2


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    Hi emira! :smile:

    Just start "given ε, ∃ n such that …" :smile:
  4. Sep 28, 2008 #3


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    Part of that, at least, is easy: If {bn} converges, then its limit is easy to get:
    [tex]\lim{n\rightarrow\infty}\frac{a_n+ a_{n+1}}{2}= \frac{\lim_{n\rightarrow\infty} a_n}{2}+ \frac{\lim_{n\rightarrow\infty} a_{n+1}}{2}[/tex]
    and those last two limits are given.

    As for proving that it does converge, since we know {an} converges to A, given any [itex]\epsilon> 0[/itex], there exist N such that if n> N, [itex]|a_n- A|< \epsilon[/itex]. Of course, if n> N, so is n+1 so the same is true of an+1!
  5. Sep 28, 2008 #4


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    \lim_{n \to \infty} a_n = A

    then it is also true that

    \lim_{n \to \infty} a_{n=1} = A

    The proof that [tex] \{b_n\}[/tex] converges is simply an exercise in the algebra of limits - no need for an [tex] \epsilon, \delta [/tex] proof (unless the original question required such).
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