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Math Help - Proving a Sequence is Cauchy proof Help!

  1. #1
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    Find Limit of a Function

    Let s_{1}=\sqrt{p}, where p>0 and

    s_{n+1}=\sqrt{p+s_n} <br />
for n in naturals
    Find the limit of  s_{n} Hint: One upper bound of the sequence is 1+2\sqrt{p}
    Last edited by fizzle45; March 21st 2010 at 07:39 PM.
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  2. #2
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    Quote Originally Posted by fizzle45 View Post
    Hi, I've been scratching my brain trying to figure out this problem.

    Show by using the definition that the following sequence is cauchy
    t_{n}=1+ \frac{1}{2!} +...+ \frac{1}{n!} for n belonging to naturals.

    I'm fairly new at Latex so I'm going to skip trying to re type all my steps with it but basically i have
    |t_{m}-t_{n}|=\frac{1}{n+1!}+\frac{1}{n+2!}+\frac{1}{n+3!  }+...+\frac{1}{m-1!}+\frac{1}{m!}

    Here's where I get a little stuck. I know that my goal is to find a bound for this which will be less than \epsilon for some N. I used a calculator and it appeared that a bound was \frac{e-1}{n!}. But I know i just can't claim something is a bound by basis of calculator values, so I need some desperate help on trying to find a bound for this thing. thanks in advance

    EDIT:
    I just read that 1+\frac{1}{1!}+\frac{1}{2!}+...\frac{1}{n!} = e if this is in fact true does anyone know where I can find the proof of this?
    The trick on these sorts of questions is to consider t_{n+1}-t_n and then use the fact that  t_{n+m}-t_n=\sum_{i=n}^{n+m-1}t_{i+1}-t_i.

    As for your edit, the limit they tend to is e. Take a look at it's Taylor expansion.
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  3. #3
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    Quote Originally Posted by fizzle45 View Post
    Hi, I've been scratching my brain trying to figure out this problem.

    Show by using the definition that the following sequence is cauchy
    t_{n}=1+ \frac{1}{2!} +...+ \frac{1}{n!} for n belonging to naturals.

    I'm fairly new at Latex so I'm going to skip trying to re type all my steps with it but basically i have
    |t_{m}-t_{n}|=\frac{1}{n+1!}+\frac{1}{n+2!}+\frac{1}{n+3!  }+...+\frac{1}{m-1!}+\frac{1}{m!}

    Here's where I get a little stuck. I know that my goal is to find a bound for this which will be less than \epsilon for some N. I used a calculator and it appeared that a bound was \frac{e-1}{n!}. But I know i just can't claim something is a bound by basis of calculator values, so I need some desperate help on trying to find a bound for this thing. thanks in advance

    EDIT:
    I just read that 1+\frac{1}{1!}+\frac{1}{2!}+...\frac{1}{n!} = e if this is in fact true does anyone know where I can find the proof of this?

    The ACTUAL fact us that \sum^\infty_{n=0}\frac{1}{n!}=e ...the sum is infinite, and either one defines the number e that way, or else defines some other way and proves the above sum is e. Anyway, any decent calculus book must include a discussion on this. For example, Apostol's "Calculus I" has several interesting parts on the number e, including its definition as that number x for which \ln x = 1 .

    Tonio
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by fizzle45 View Post
    Let s_{1}=\sqrt{p}, where p>0 and

    s_{n+1}=\sqrt{p+s_n} <br />
for n in naturals
    Find the limit of  s_{n} Hint: One upper bound of the sequence is 1+2\sqrt{p}
    Let  L be the limit of the sequence  s_n .

    Since  s_{n}^2=p+s_{n-1} and for large enough N\; |s_n-s_{n-1}|<\epsilon ,  L^2=p+L .

    Hence  L is the positive solution to  x^2-x-p=0 .

    Thus  L = \frac{1+\sqrt{4p+1}}{2} .
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  5. #5
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    Cauchy sequence

    Quote Originally Posted by tonio View Post
    The ACTUAL fact us that \sum^\infty_{n=0}\frac{1}{n!}=e ...the sum is infinite, and either one defines the number e that way, or else defines some other way and proves the above sum is e. Anyway, any decent calculus book must include a discussion on this. For example, Apostol's "Calculus I" has several interesting parts on the number e, including its definition as that number x for which \ln x = 1 .

    Tonio
    As far as the original problem of showing that \sum_{k=0}^n \frac{1}{n!} is Cauchy, you could try to use
    n!\ge 2^{n-1}
    \frac1{n!}\le \frac1{2^{n-1}}
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