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Thread: Limit of sequence

  1. #1
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    Limit of sequence

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

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

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

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

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

    According to what you wrote:

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

    $\displaystyle f_2(n)=\left(\frac{f_1(n)}{\lim\limits_{p\to\infty }f_1(p)}\right)^n=
    \left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right )^n=f_1(n)$ 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:

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

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

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

    So your assumption is not true.

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


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


    So your assumption is not true.

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


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

    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:

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

    and that's it.

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

    Let $\displaystyle k\geq 1$. Start from:

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

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

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

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

    (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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