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!

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




Similar Discussions: Convergence of a sequence
  1. Sequence Convergence (Replies: 6)

  2. Sequence Convergence (Replies: 2)

Loading...