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Math Help - Limit of sequence

  1. #1
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    Limit of sequence

    Consider the sequence f_{0}:N^{*}->R
    f_{0}(n)=(1+\frac{1}{n})^{n}
    and define,
    f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}.
    n,k positive natural numbers.
    Find:
    \lim_{n\to\infty}f_{k}(n)

    Using induction I found that
    f_{k}(n)=\frac{(1+\frac{1}{n})^{n^{k+1}}}{e^{\sum_  {i=1}^{k}(-1)^{i+1}\frac{n^{k+i-1}}{i}}}
    But I don't know how to find the limit.

    The limit should be e^{\frac{(-1)^{k}}{k+1}}
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  2. #2
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    Quote Originally Posted by m3th0dman View Post
    Consider the sequence f_{0}:N^{*}->R
    f_{0}(n)=(1+\frac{1}{n})^{n}
    and define,
    f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}.
    n,k positive natural numbers.
    Find:
    \lim_{n\to\infty}f_{k}(n)

    Using induction I found that
    f_{k}(n)=\frac{(1+\frac{1}{n})^{n^{k+1}}}{e^{\sum_  {i=1}^{k}(-1)^{i+1}\frac{n^{k+i-1}}{i}}}
    But I don't know how to find the limit.

    The limit should be e^{\frac{(-1)^{k}}{k+1}}

    According to what you wrote:

    f_1(n)=\left(\frac{f_0(n)}{\lim\limits_{n\to\infty  }f_0(n)}\right)^n=\left(\frac{\left(1+\frac{1}{n}\  right)^n}{e}\right)^n \xrightarrow [n\to\infty]{}1 , so:

    f_2(n)=\left(\frac{f_1(n)}{\lim\limits_{p\to\infty  }f_1(p)}\right)^n=<br />
\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right  )^n=f_1(n) and etc...., so I think you may have not written what you meant...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    According to what you wrote:

    f_1(n)=\left(\frac{f_0(n)}{\lim\limits_{n\to\infty  }f_0(n)}\right)^n=\left(\frac{\left(1+\frac{1}{n}\  right)^n}{e}\right)^n \xrightarrow [n\to\infty]{}1 , so:

    f_2(n)=\left(\frac{f_1(n)}{\lim\limits_{p\to\infty  }f_1(p)}\right)^n=<br />
\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right  )^n=f_1(n) and etc...., so I think you may have not written what you meant...

    Tonio
    The following proposition is false:
    \lim_{n\to\infty}1^{n}=1

    So your assumption is not true.

    \lim_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{e})^n=e  ^{-\frac{1}{2}}
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  4. #4
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    Quote Originally Posted by m3th0dman View Post
    The following proposition is false:
    \lim_{n\to\infty}1^{n}=1


    In fact this is trivially true. What is not true in general is that f(n)\xrightarrow [n\to\infty]{}1\Longrightarrow f(n)^n\xrightarrow [n\to\infty]{}1


    So your assumption is not true.

    No assumption but sheer misreading and consequent mistake of mine...


    \lim_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{e})^n=e  ^{-\frac{1}{2}}

    Indeed. And this limit is pretty hard to evaluate: I had to use twice L'Hospital' rule with logarithmic notation...

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    Indeed. And this limit is pretty hard to evaluate: I had to use twice L'Hospital' rule with logarithmic notation...

    Tonio
    I have been taught to avoid L'Hospital' rule, so here is what I wrote:

    \left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right  )^n = e^{n(n\log(1+\frac{1}{n})-1)} =e^{n(n(\frac{1}{n}-\frac{1}{2n^2}+o(\frac{1}{n^2}))-1)}=e^{-\frac{1}{2}+o(1)},

    and that's it.

    Quote Originally Posted by m3th0dman View Post
    Consider the sequence f_{0}:N^{*}->R
    f_{0}(n)=(1+\frac{1}{n})^{n}
    and define,
    f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}.
    n,k positive natural numbers.
    Find:
    \lim_{n\to\infty}f_{k}(n)
    Moreover, the same method works for this general problem. Let us write g_k(n)=\log f_k(n) for simplicity, and \ell_k=\lim_n g_k(n), for all k.

    Let k\geq 1. Start from:

    \log\left(1+\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{2n^2}+\cdots+\frac{(-1)^{k+1}}{k n^k}+o\left(\frac{1}{n^k}\right).

    The definition of \ell_k is just an intricate way to describe the coefficients in this expansion.

    Indeed,
    g_0(n)=n\log(1+\frac{1}{n})=1-\frac{1}{2n}+\cdots+\frac{(-1)^{k+1}}{k n^{k-1}}+o\left(\frac{1}{n^{k-1}}\right)=1+o(1),
    hence \ell_0=1,
    g_1(n)=n(g_0(n)-\ell_0)=n(g_0(n)-1)=-\frac{1}{2}+\frac{1}{3n}+\cdots+\frac{(-1)^{k+1}}{k n^{k-2}}+o\left(\frac{1}{n^{k-2}}\right)=-\frac{1}{2}+o(1),
    hence \ell_1=-\frac{1}{2},
    and so on, "unfolding the expansion", until
    g_{k-1}(n)=n(g_{k-2}(n)-\ell_{k-2})=\frac{(-1)^{k+1}}{k}+o(1),
    hence \ell_{k-1}=\frac{(-1)^{k+1}}{k}.

    If you really want to write every word of the proof, then you have to explicitate the above induction. Let k\geq 1 and prove by induction on j=0,\ldots,k-1 that :
    g_j(n)=\frac{(-1)^{j}}{j+1}+\frac{(-1)^{j+1}}{j+2}\frac{1}{n}+\cdots+\frac{(-1)^{k+1}}{k}\frac{1}{n^{k-j-1}}+o\left(\frac{1}{n^{k-j-1}}\right)
    (hence \ell_j=\frac{(-1)^{j}}{j+1}).

    (you could also use convergent series instead of asymptotic expansions to shorten the proof, but these would be a little inappropriate here since we only care for the limit)
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  6. #6
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    Thanks

    Thank you very much!
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