1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    Hi emira! :smile:

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Homework Helper


    \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).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Advanced Calculus Proof-limit of a sequence.