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Math Help - prove that

  1. #1
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    prove that

    x^n/n! = 0 for n-> 0
    using sandwich theorem

    a sequence given by a(n+1) = SQRT(2 + an)
    prove that it is converging.. I found the limit is 2.. but how to prove it converges?

    prove that if an converges to a and bn converges to b then an/bn converges to a/b using the theorem |an-l| < E for all n > N and for any E > 0
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: prove that

    Quote Originally Posted by ice_syncer View Post
    ... a sequence given by a(n+1) = SQRT(2 + an)... prove that it is converging.. I found the limit is 2.. but how to prove it converges?...
    The recursive relation can be written as...

    \Delta_{n}= a_{n+1}-a_{n} = \sqrt{2+a_{n}}-a_{n}= f(a_{n})

    The function f(x) is represented here...



    There is only one 'attractive fixed point' [i.e. a point where f(x) crosses the x axis with negative slope...] at x_{0}=2 and, because for x>-2 is |f(x)|<|x_{0}-x| [see the 'red line'...] any a_{0}>-2 will produce a sequence converging to 2 monotonically...

    Kind regards

    \chi \sigma
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    Re: prove that

    How can you show it without graph ?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: prove that

    If in general we write the 'recursive relation' as...

     \Delta_{n}= a_{n+1}-a_{n}= f(a_{n}) (1)

    ... then is...

    a_{n}= a_{0}+ \sum_{k=0}^{n-1} \Delta_{k} (2)

    ... and the convergence of the series in (2) implies the convergence of the a_{n}. If we suppose that f(x) is continous and vor all n is \Delta_{n}>0 or \Delta_{n}<0, then the series in (2) converges if...

    a) \lim_{n \rightarrow \infty} \Delta_{n}=0...

    b) for n 'large enough' is \frac{\Delta_{n+1}}{\Delta_{n}} \le \lambda < 1 [ratio test]...

    The condition a) is satisfied if it exist a real number x_{0} for which is f(x_{0}=0), and in such case is [if it exists...] \lim_{n \rightarrow \infty} a_{n}= x_{0}. The condition b) is satisfied if is f^{'}(x_{0}) <0 and there exist an interval around x_{0} where is |f(x)|< |x_{0}-x|...

    Kind regards

    \chi \sigma
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  5. #5
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    Re: prove that

    lim(n->inf)(x^n/n!)
    case 1: |x|<=1
    => 0<=|x^n/n!|<=|1/n!| for all n>=1 => lim(n->inf)(x^n/n!)=0 by the squeeze thm for all |x|<1

    case 2: |x|>1 ;
    let c(x)=Product(for i from 1 to ceiling(|x|) of x/i) is constant w.r.t n
    let d(x)=|x|/(ceiling(|x|)+1) =>0<d(x)<1
    lim(n->inf)(d(x)^n)=0, |x/i|<=d(x) for all i>ceiling(|x|)

    0<=|x^n/n!|=|c(x)||Product(for i from (ceiling(|x|)+1) to n of x/i)|
    <=|c(x)|(Product for i from (ceiling(|x|)+1) to n of d(x))
    =|c(x)|d(x)^n. for all natural numbers n

    hence by the squeeze thm x^n/n! ->0 for |x|>1

    Thus for all real numbers x, x^n/n! ->0 for n->inf

    The main idea i think is to write the nth term as a product over i of x/i and to consider the first cieling(x) terms as a constant. each consective term is x/n times the previous so if it were to converge to L, L=lim(n->inf)(x^n/n!)=lim(n->inf)(x/n)*lim(n->inf)(x^(n-1)/(n-1)!)=0*L
    so L must be 0.
    Last edited by JeffN12345; August 24th 2011 at 01:06 PM.
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  6. #6
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    Re: prove that

    Quote Originally Posted by ice_syncer View Post
    prove that if an converges to a and bn converges to b then an/bn converges to a/b using the theorem |an-l| < E for all n > N and for any E > 0
    first note: b must be non zero for a/b to exist so we require b nonzero
    second note: b is nonzero and b_n->b => S={k:b_k=0} is finite so let n_0=max(S).
    we begin by deleting the first n_0 terms from both the sequences {a_n} and {b_n} (this will not affect convergence as we do not alter the 'tails' of the sequences) and adjusting the indices so they both start at 1. This is in order to ensure {a_n/b_n} is a sequence of real numbers. From now on I assume b_n is non zero for all n.

    b_n converges to a real number b and a is a real number
    => there exists max{(b_n)}, max{a/(b*b_n)} in real numbers

    let c=max{(b_n)}, d=max{a/(b*b_n)}

    i) {a_n}->a => for all E>0 there exists N s.t. |a_n-a|<E*c/2 for all n>=N
    ii) {b_n}->b => for all E>0 there exists M s.t. |b_n-b|<E*d/2 for all n>=M

    0<=|a_n/b_n-a/b|
    =|(b*a_n)-(a_n*b)|/|b_n*b|
    =|(b*a_n-b*a)+(b*a-b_n*a)|/|b_n*b|
    <=(|(b*a_n-b*a)|+|(b*a-b_n*a)|)/|b_n*b|
    =(|a_n-a|/|b_n|)+(|a/(b*b_n)||b-b_n|)
    <=E/2+E/2=E for n>= max(M,N)

    hence a_n/b_n->a/b if n->inf
    Last edited by JeffN12345; August 24th 2011 at 02:43 PM.
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  7. #7
    Super Member redsoxfan325's Avatar
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    Re: prove that

    Quote Originally Posted by ice_syncer View Post
    x^n/n! = 0 for n-> 0
    using sandwich theorem
    Let a=\lceil |x|\rceil

    |x^n|\leq a^n=a^aa^{n-a}\leq a^a\frac{(n-1)!}{(a-1)!}

    (The last inequality holds because there are n-a terms in \frac{(n-1)!}{(a-1)!}, all of which are \geq a.)

    Given this,

    0\leq\left|\frac{x^n}{n!}\right|\leq\frac{a^a}{(a-1)!}\frac{(n-1)!}{n!}=\frac{a^a}{(a-1)!}\cdot\frac{1}{n}

    Hence the squeeze theorem says the limit is 0.

    An easier way would be to use the ratio test:

    \lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{(n+1)!  }}{\frac{x^n}{n!}}\right|=\lim_{n\to\infty}\left| \frac{x}{n+1}\right|=0

    Therefore the sum converges so the terms must go to zero.
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  8. #8
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    Re: prove that

    Isn't it suffiencient to show that it is increasing and has a limit? How do you show that the recursive sequence is increasing ?
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  9. #9
    Super Member redsoxfan325's Avatar
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    Re: prove that

    Quote Originally Posted by ice_syncer View Post
    Isn't it suffiencient to show that it is increasing and has a limit? How do you show that the recursive sequence is increasing ?
    There are a few cases to consider:

    1. Suppose a_n > 2. Then \sqrt{a_n+2} > \sqrt{2+2}=2. Furthermore, since both sides are positive, a_n > \sqrt{a_n+2} iff a_n^2>a_n+2 iff a_n^2-a_2-2=(a_n-2)(a_n+1)>0. Since a_n > 2 by assumption, the sequence is monotone decreasing and bounded below by 2. So if a_0 > 2, the sequence is monotone decreasing and converges to 2.

    2. Suppose 0\leq a_n < 2. Then \sqrt{a_n+2} < \sqrt{2+2}=2. The same reasoning as before shows that if 0\leq a_0 < 2, the sequence is monotone increasing and converges to 2.

    3. Suppose -2\leq a_n<0. Then \sqrt{a_n+2} < \sqrt{2+2}=2 as before, and a_n < \sqrt{a_n+2} because the LHS is negative and the RHS is positive, so the sequence is monotone increasing.

    4. If a_0 < -2, this isn't a real sequence anymore, so we may ignore this case.

    The conclusion is that the sequence converges to 2 in the following manner:
    -2 \leq a_0 < 2\implies\mbox{monotone~increasing}
    a_0=2\implies\mbox{stationary}
    a_0 > 2\implies\mbox{monotone~decreasing}
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