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Homework Help: Show sequence is cauchy

  1. Oct 16, 2008 #1
    For n in the naturals, let

    [tex]p_n = 1 + \frac{1}{2!} + ... + \frac{1}{n!}[/tex]

    Show it is cauchy.


    I have set up |p_n+k - p_n | < e , and I have solved for this.

    I got [tex]|p_{n+k} - p_n | = \frac{1}{(n+1)!} + ... + \frac{1}{(n+k)!}[/tex]

    I am trying to follow an example in the book. I now need to find a telescoping sequence that is a little bit greater than my sequence above. It should also contain what seems like two fractions, or parts, with one of them converging and the other adding an arbitrary "k". I would appreciate any hints on this.

    I got stuck on how to break up, 1 / (n+k-1)! . I am not sure if that is worthwhile or I am totally off.
  2. jcsd
  3. Oct 16, 2008 #2


    Staff: Mentor

    You have k terms, the largest of which is 1/(n + 1)! Is that enough of a hint?
  4. Oct 16, 2008 #3
    Your hint helped but since this is my first cauchy problem I am still lost. I am getting lost when I try to find the partial fractions of


    I am doing this in order to get the telescoping sequence.
  5. Oct 16, 2008 #4


    Staff: Mentor

    So why do you think you need to decompose the fraction and why do you think you need a telescoping sequence? Given a positive number epsilon, all you need to do is find a number N so that for all m and n larger than N, any two terms in your sequence are closer together than epsilon.

    HINT: You have k terms (count 'em!) on the right, the largest of which is 1/(n + 1)!
  6. Oct 17, 2008 #5
    I'm not sure how to use your hint, Mark44....
    are you going to prove it by saying that
    which looks as if it approaches to 0 as N goes to infinity. So it can be made less than a given epsilon?
    But, I realize that k is independent on the given epsilon. So if you find a N. I can choose k large enough, say, k=(n+1)! which will make that inequality useless.
    So how can it be done?

    A different hint: prove its convergenece as an upper bounded increasing sequences, which implies also it's a cauchy.

    PS. Well, after a second thought, The hint of Mark44 really works, but a bit tricky:surprised
    Last edited: Oct 17, 2008
  7. Oct 17, 2008 #6
    I got [tex] \frac{k}{(n+1)!} = \frac{k}{n!} - \frac{kn}{(n+1)!} [/tex]

    I can't get anywhere. I want to say something like k/n! is convergent, therefore it is cauchy and we can say that,

    [tex]| \frac{k}{n!} - \frac{kn}{(n+1)!} | < e [/tex]

    I know that can't work because there is an n in the numerator but what else can I do?
  8. Oct 17, 2008 #7
    Okay, I feel good. Somebody shoot me down!

    Since 1/n! converges to 0 and n->inf , we can let e>0 and say there exists N in Naturals such that for all n >= N,

    | 1/n! - 0 | < e/k.


    | s_n+k - s_n | = 1/(n+1)! + ... + 1/(n+k)! < k/ (n+1)! < k/n! < k(e/k) = e.

    Please, please, please let this be right. I am ready to move on.
  9. Oct 17, 2008 #8
    this is not right.. as I said before, your k should be independent on epsilon.
    [tex]\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1}{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{k}})<\frac{1}{n!n}[/tex]
    so we can choose a N large enough, no matter how large your k is, its sum is less than a given epislon.
  10. Oct 18, 2008 #9
    \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<

    [tex] < \frac{1}{(n+1)!} \sum_{j=0}^{\infty} \frac{1}{(n+1)^j} < \sum_{j=0}^{\infty} \frac{1}{(n+1)^j} = \frac{n+1}{n} [/tex]
    Last edited: Oct 18, 2008
  11. Oct 20, 2008 #10
    no...it is useless that you have "........< (n+1)/n"
  12. Oct 20, 2008 #11
    Choose N> 1/e. Then for all n > N we have that,


    \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1}{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{k}})<\frac{1}{n!n}<\frac{1}{n}<\varepsilon


    Therefore, [tex]|s_{n+k}-s_n|<\varepsilon[/tex].

    Is that right?

    EDIT: Is that LaTeX showing up for anybody else?
    Last edited: Oct 20, 2008
  13. Oct 20, 2008 #12
    \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<\frac{1}{n!n}<\frac{1}{n}<\varepsilon

    \\Edit: indeed latex isn't working
  14. Oct 20, 2008 #13
    [tex]\frac{1}{n!} \int^{1}_{0} sin x dx[/tex]

    Edit: indeed
  15. Oct 22, 2008 #14

    \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...+\frac{1}{(n+ k)!}=\frac{1}{(n+1)!} (1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\frac{1 }{(n+2)...(n+k)})<\frac{1}{(n+1)!} (1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\frac{1}{(n +1)^{k}})<\frac{1}{n!n} [/tex]



    Latex is working again :cool:
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