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Math Help - Sequences... help!

  1. #1
    Rui
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    Sequences... help!

    Consider the sequence {an} = {n!/nⁿ}
    a) Write n! = 1*2*...*n-1*n and nⁿ = n*n *....*n*n to express an as a product of n terms.
    b) Show that 0≤ an≤ 1/n and use the Squeeze Theorem to show that the sequence converges.
    c) Show that the sequence {cⁿ/n!} converges for c a positive constant.
    d) Show that the sequence {nⁿ/((2n)!)} converges.

    I don't get any of this help!
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Rui View Post
    Consider the sequence {an} = {n!/nⁿ}
    a) Write n! = 1*2*...*n-1*n and nⁿ = n*n *....*n*n to express an as a product of n terms.
    \frac{n!}{n^n}=\frac{{\color{red}1}*2*{\color{blue  }3}*\dots*{\color{magenta}n}}{{\color{red}n}*n*{\c  olor{blue}n}\dots*{\color{magenta}n}}= \underbrace{{\color{red}\frac 1n} * \frac 2n * {\color{blue}\frac 3n} * \dots * {\color{magenta}\frac nn}}_{\text{n terms}}


    b) Show that 0≤ an≤ 1/n and use the Squeeze Theorem to show that the sequence converges.
    You can easily show it's superior to 0.
    Now, you see that the first term is 1/n.
    All the terms following 1/n are inferior to 1.

    a_n=\frac 1n * \underbrace{\frac 2n}_{<1} * \underbrace{\frac 3n}_{<1} * \dots * \underbrace{\frac nn}_{\le 1}

    Hence a_n \le \frac 1n

    And because it's bounded by sequences which converge, this converges.
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  3. #3
    Rui
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    Thanks so much Moo, I kinda get it now. But I still don't have any idea on how to do c) and d). Any help would be great.
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  4. #4
    Moo
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    Hm... I didn't see it was for College/University, so I was a bit lost

    The same way, we write this as a product with n terms.

    \frac{c^n}{n!}=\frac c1 * \frac c2 * \frac c3 * \dots * \frac cn

    Since we work while n tends to infinity, it'll be useful to say that there exists N such that for all n>N, n>c.

    \lim_{n \to \infty} \frac{c^n}{n!}=\underbrace{\frac c1 * \frac c2 * \frac c3 * \dots * \frac c {N-1}}_{\color{red}\text{finite number of terms } \rightarrow \text{ finite value=X}} * \underbrace{\frac cN}_{<1} * \underbrace{\frac c {N+1}}_{<1} * \dots

    So 0 \le \lim_{n \to \infty} \frac{c^n}{n!} \le {\color{red}X}


    --> The sequence converges.

    I think it's correct, but I have a very slight hesitation..
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Rui View Post
    Consider the sequence {an} = {n!/nⁿ}
    a) Write n! = 1*2*...*n-1*n and nⁿ = n*n *....*n*n to express an as a product of n terms.
    b) Show that 0≤ an≤ 1/n and use the Squeeze Theorem to show that the sequence converges.
    c) Show that the sequence {cⁿ/n!} converges for c a positive constant.
    d) Show that the sequence {nⁿ/((2n)!)} converges.

    I don't get any of this help!
    For part D I see the most enlightening answer comes from seeing that

    \forall{n}>0,n^n\leq(n!)^2

    and that \forall{n}>0,\frac{a^n}{n^n}\leq\frac{n^n}{(2n)!}


    So then taking the limits does not change the inequality so coming from the above statement we can see that

    \lim_{n\to\infty}\frac{a^n}{(2n)!}\leq\lim_{n\to\i  nfty}\frac{n^n}{(2n)!}\leq\lim_{n\to\infty}\frac{(  n!)^2}{(2n)!}

    Now we consider the limits as n\to\infty for each of these functions

    The limit of the left side obviously goes to zero considering its an exponential function over an exponential function and a\ll{n} as n\to\infty


    Alternatively for the rigth side you could Let L=\lim_{n\to\infty}\frac{a^n}{n^n}=\lim_{n\to\inft  y}\bigg(\frac{a}{n}\bigg)^n

    Now taking the natural log of both sides we get

    \ln(L)=\lim_{n\to\infty}\ln\bigg[\bigg(\frac{a}{n}\bigg)^n\bigg]=\lim_{n\to\infty}n\ln\bigg(\frac{a}{n}\bigg)

    Now looking at this we see that direct substitution gives

    \ln(L)=\infty\cdot{-\infty}=-\infty

    NOTE This product yielding -∞ may be too informal for your school. At mine it would be excepted

    So we have that \ln(L)=-\infty\Rightarrow{L=e^{-\infty}=0}

    Thus \lim_{n\to\infty}\frac{a^n}{n^n}=0




    Now we consider \lim_{n\to\infty}\frac{(n!)^2}{(2n)!}

    There are many advanced ways to do this but the most intutive and easiest to show way is to do this

    Expanding we get


    \lim_{n\to\infty}\frac{\bigg(1\cdot{2}\cdot{3}\cdo  t...\cdot{(n-1)}\cdot{(n)}\bigg)\bigg(1\cdot{2}\cdot{3}\cdot...  \cdot{(n-1)}\cdot{(n)}\bigg)}{1\cdot{2}\cdot{3}\cdot...\cdo  t{(n-1)}\cdot{n}\cdot...\cdot{(2n-1)}\cdot{(2n)}}

    Cancelling we get

    \lim_{n\to\infty}\frac{1\cdot{2}\cdot{3}\cdot....\  cdot{(n-1)}\cdot{(n)}}{(n+1)\cdot{(n+2)}\cdot{(n+3)}\cdot.  ..\cdot{(2n-1)}\cdot{(2n)}}


    Now it is clear to see that the denominator as n\to\infty is infinitely greater than the numerator thus making

    \lim_{n\to\infty}\frac{(n!)^2}{(2n)!}=0



    So now we look back to our original problem


    \lim_{n\to\infty}\frac{a^n}{n^n}\leq\lim_{n\to\inf  ty}\frac{n^n}{(2n)!}\leq\lim_{n\to\infty}\frac{(n!  )^2}{(2n)!}



    So then seeing as we showed that \lim_{n\to\infty}\frac{a^n}{n^n}=0

    and \lim_{n\to\infty}\frac{(n!)^2}{(2n)!}=0

    We can rewrite this inequality to read

    0\leq\lim_{n\to\infty}\frac{n^n}{(2n)!}\leq{0}

    Then by the squeeze theorem we conclude that

    \lim_{n\to\infty}\frac{n^n}{(2n)!}=0


    And therefore is convergent
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  6. #6
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    For part D I see the most enlightening answer comes from seeing that

    \forall{n}>0,n^n\leq(n!)^2

    and that \forall{n}>0,\frac{a^n}{n^n}\leq\frac{n^n}{(2n)!}
    And where does it come from ?
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