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Homework Help: Convergence of a sequence

  1. Dec 9, 2012 #1
    I think the solution I've found makes sense, but I'd like it to be double-checked.
    1. The problem statement, all variables and given/known data

    Let ##(a_n)## be a limited sequence and ##(b_n)## such that ##0≤b_n≤ \frac{1}{2} B_{n-1} ##

    Prove that if
    ##a_{n+1} \ge a_{n} -b_{n}##

    Then
    ##\lim_{n\to \infty}a_n##

    exists.

    3. The attempt at a solution

    I can say that ##b_n \ge (\frac{1}{2})^n b_0 ## which is constant.
    THen,
    ##a_n-a_{n+1} \ge (\frac{1}{2})^nb_0##

    Thus ##|a_n-a{n+1}| ## is a Cauchy sequence, which means it converges and therefore the limit exists.
     
  2. jcsd
  3. Dec 9, 2012 #2
    You need to watch your inequality signs a bit. From ##0\leq b_n\leq\frac12b_{n-1}## you get ##0\leq b_n\leq(\frac12)^nb_0##. And that isn't a constant, since it depends on ##n##. What you probably mean is that ##b_0## is constant.

    Then, ##a_n-a_{n+1}\leq(\frac12)^nb_0##. But how do you now conclude that ##(|a_n-a_{n+1}|)_{n\in\mathbb N}## is a Cauchy sequence? And even if it is, how does that make the sequence ##(a_n)_{n\in\mathbb N}## converge?
     
  4. Dec 9, 2012 #3
    Right, ##\frac{1}{2}^nb_0## is not a constant, but for n diverging to ##\infty b_0## gets smaller and smaller, so ##a_n-a_{n+1}## is bounded above.
    For Cauchy's criterion a sequence converges to something if and only if this holds:
    for every  ##\epsilon> 0## there exists n* such that ##|a_n-a_m| < \epsilon ## whenever n, m > n*.
    Let's just call ##\frac{1}{2}b_0=\epsilon## and ##a_m=a_{n+1}##. for a n*, big enough, the criterion holds.
     
  5. Dec 9, 2012 #4

    pasmith

    User Avatar
    Homework Helper

    That's not sufficient.

    You need to show that for all [itex]\epsilon > 0[/itex], there exists [itex]N \in \mathbb{N}[/itex] such that for all [itex]n,m \geq N[/itex], [itex]|a_n - a_m| < \epsilon[/itex]. You can't just choose a particular [itex]\epsilon[/itex].

    Hint: Without loss of generality you can take [itex]m > n[/itex]. Then
    [tex]|a_n - a_m| = |a_n - a_{n+1} + a_{n+1} + \dots - a_{m-1} + a_{m-1} - a_m|
    [/tex]
     
    Last edited: Dec 9, 2012
  6. Dec 9, 2012 #5
    ok, what if I prove it this way:
    by definition, a sequent is convergent when the following property holds:
    for every ## \epsilon>0 ## there exists an ##N## such that for every n## \geq N##, we have
    ## |a_n-L|<\epsilon##.
    In this exercise I have ##|a_n - a_m| < \epsilon##
    but by taking
    ##n,m \geq N##
    I can obtain
    ##|a_n-a_m+L-L|<\epsilon ##which is ## \geq |a_n-L|+|a-M-L| \geq \frac{\epsilon}{2}+\frac{\epsilon}{2} ##
    is this what you meant?
     
  7. Dec 9, 2012 #6
    ##b_0## doesn't get smaller, ever. You really should be more careful about what you put in words, because you need even more care with formulas.

    You don't have ##|a_n-a_m|<\epsilon##. That's what you want to get. What you're using here is called "circular reasoning", and it's a very easy mistake to make -- confusing what you are aiming at with what you're starting off with.

    I think you might try pasmith's suggestion: Assume ##m>n## and have a look at
    $$
    |a_n-a_m| = |a_n-a_{n-1}+a_{n-1}-a_{n-2}+a_{n-2}-\ldots-a_{m+2}+a_{m+2}-a_{m+1}+a_{m+1}-a_m|
    $$
     
  8. Dec 11, 2012 #7
    thanks! is it okay now?:
    ##|a_n - a_m| \le |a_n - a_{n-1}| + |a_{n-1}-a_{n-2}|+...+|a_m - a_{m-1}| ##
    each member of this sum is ≤ ##(\frac{1}{2})^nb_0## thus ##|a_n-a_m|## converges.
    (i don't know exactly how to formally find ##\epsilon##
    can I also add that ##b_n## goes to 0 due to the squeeze rule?
    thanks again
     
  9. Dec 11, 2012 #8
    You must check your inequalities better. Each member of that sum is ##\leq(\frac12)^{n-1}b_0## (note the exponent).

    Also, you don't find ##\epsilon##. That value is just assumed to be positive, and you must show that the sum is smaller than ##\epsilon##. So you need to ensure that
    $$
    \left(\frac12\right)^{n-1}b_0 < \epsilon.
    $$
    Suppose you solve that and get some integer ##N## as a result. Then you can re-write your proof "backwards":

    Let ##\epsilon>0## be given. Then we set ##N=(\ldots \epsilon \ldots)##, and for ##n,m\geq N## with ##m>n## we have
    $$
    \epsilon > \left(\frac12\right)^Nb_0 \geq \ldots \geq |a_n-a_m|.
    $$
    This proves that ##(a_n)## is a Cauchy sequence, and so this sequence must converge. QED
     
  10. Dec 11, 2012 #9
    thank you, i really didn't get it
     
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