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Thread: 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.
    $\displaystyle \frac{n!}{n^n}=\frac{{\color{red}1}*2*{\color{blue }3}*\dots*{\color{magenta}n}}{{\color{red}n}*n*{\c olor{blue}n}\dots*{\color{magenta}n}}= \underbrace{{\color{red}\frac 1n} * \frac 2n * {\color{blue}\frac 3n} * \dots * {\color{magenta}\frac nn}}_{\text{n terms}}$


    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.

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

    Hence $\displaystyle a_n \le \frac 1n$

    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.

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

    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.

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

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


    --> 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

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

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


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

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

    Now we consider the limits as $\displaystyle n\to\infty$ 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 $\displaystyle a\ll{n}$ as $\displaystyle n\to\infty$


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

    Now taking the natural log of both sides we get

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

    Now looking at this we see that direct substitution gives

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

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

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

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




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

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

    Expanding we get


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

    Cancelling we get

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


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

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



    So now we look back to our original problem


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



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

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

    We can rewrite this inequality to read

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

    Then by the squeeze theorem we conclude that

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


    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

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

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