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Thread: Problem 39

  1. #1
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    Problem 39

    Let $\displaystyle a,b\in \mathbb{Z}^+$. For each positive integer $\displaystyle n$ let $\displaystyle H_n(a,b) = \frac{1}{a+b}+\frac{1}{2a+b}+...+\frac{1}{na+b}$.
    Find the limit,
    $\displaystyle \lim \ \frac{H_n(a,b)}{H_n(c,d)}$

    (Where $\displaystyle c,d$ are possibly different integers defining a different sequence).
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let $\displaystyle a,b\in \mathbb{Z}^+$. For each positive integer $\displaystyle n$ let $\displaystyle H_n(a,b) = \frac{1}{a+b}+\frac{1}{2a+b}+...+\frac{1}{na+b}$.
    Find the limit,
    $\displaystyle \lim \ \frac{H_n(a,b)}{H_n(c,d)}$

    (Where $\displaystyle c,d$ are possibly different integers defining a different sequence).
    is this lim as n approaches infinity?
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    Quote Originally Posted by kalagota View Post
    is this lim as n approaches infinity?
    When you are dealing with sequences that is the only type of limit you can have. (What else can you possibly approach )
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    Maybe I'm off base here, PH,:

    $\displaystyle \lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\fr ac{1}{na+b}\right)=\frac{1}{a}$

    $\displaystyle \lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\fr ac{1}{nc+d}\right)=\frac{1}{c}$

    So, we have: $\displaystyle \frac{\frac{1}{a}}{\frac{1}{c}}=\boxed{\frac{c}{a} }$
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    Quote Originally Posted by galactus View Post
    Maybe I'm off base here, PH,:

    $\displaystyle \lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\fr ac{1}{na+b}\right)=\frac{1}{a}$

    $\displaystyle \lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\fr ac{1}{nc+d}\right)=\frac{1}{c}$
    The sums diverges. Because,
    $\displaystyle 0\leq \frac{1}{n+n}\leq \frac{1}{na+b}$ for sufficiently large $\displaystyle n$.

    And $\displaystyle \sum_{n=1}^{\infty}\frac{1}{2n}$ does not converge, it is the harmonic series.
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    MHF Contributor red_dog's Avatar
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    Let $\displaystyle x_n=H_n(a,b), \ y_n=H_n(c,d)$.
    $\displaystyle (y_n)$ is ascending and unbounded.
    By Stolz-Cesaro, we have
    $\displaystyle \displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\to\infty}\frac{\frac{1}{a(n+1)+b}}{\f rac{1}{c(n+1)+d}}=\lim_{n\to\infty}\frac{cn+c+d}{a n+a+b}=\frac{c}{a}$.

    Then $\displaystyle \displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\fra c{c}{a}$
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    Quote Originally Posted by red_dog View Post
    Let $\displaystyle x_n=H_n(a,b), \ y_n=H_n(c,d)$.
    $\displaystyle (y_n)$ is ascending and unbounded.
    By Stolz-Cesaro, we have
    $\displaystyle \displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\to\infty}\frac{\frac{1}{a(n+1)+b}}{\f rac{1}{c(n+1)+d}}=\lim_{n\to\infty}\frac{cn+c+d}{a n+a+b}=\frac{c}{a}$.

    Then $\displaystyle \displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\fra c{c}{a}$
    I can do it without Stolz-Cesaro. Want to try?
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    The key step is to note that $\displaystyle H_n(a,b)\sim \ln (an+b)$. That means, $\displaystyle \lim \ \frac{H_n(a,b)}{H_n(c,d)} = \lim \ \frac{\ln (an+b)}{\ln (cn+d)}=\frac{c}{a}$.
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