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Thread: [SOLVED] Positive and divergent infinite sequence

  1. #1
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    [SOLVED] Positive and divergent infinite sequence

    I need to construct a divergent sequence $\displaystyle \{ a_n \}$ such that $\displaystyle a_n>0, \forall n \in \mathbb{N}$, and $\displaystyle \lim \limits_{n \to \infty } \frac{a_{n+1}}{a_n}=1$

    I can't think of one
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  2. #2
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    Hello,
    Quote Originally Posted by akolman View Post
    I need to construct a divergent sequence $\displaystyle \{ a_n \}$ such that $\displaystyle a_n>0, \forall n \in \mathbb{N}$, and $\displaystyle \lim \limits_{n \to \infty } \frac{a_{n+1}}{a_n}=1$

    I can't think of one
    $\displaystyle a_n=\frac 1n$ ?


    Note that if you apply the ratio test and find the limit to be 1, the test is inconclusive.
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    Quote Originally Posted by Moo View Post
    Hello,

    $\displaystyle a_n=\frac 1n$ ?


    Note that if you apply the ratio test and find the limit to be 1, the test is inconclusive.
    The sequence $\displaystyle a_n=\frac 1n$ converges to 0, because $\displaystyle \lim \limits_{n \to \infty} a_n=\lim \limits_{n \to \infty} \frac{1}{n}=0$.

    That will work for the Harmonic series $\displaystyle \sum ^\infty _{n=1} \frac{1}{n}$, but I don't think that will work in this case.

    I need something like $\displaystyle \lim \limits_{n \to \infty} a_n=\infty$ and $\displaystyle \lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$
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  4. #4
    Moo
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    Quote Originally Posted by akolman View Post
    The sequence $\displaystyle a_n=\frac 1n$ converges to 0, because $\displaystyle \lim \limits_{n \to \infty} a_n=\lim \limits_{n \to \infty} \frac{1}{n}=0$.

    That will work for the Harmonic series $\displaystyle \sum ^\infty _{n=1} \frac{1}{n}$, but I don't think that will work in this case.
    Please excuse me, I thought we were working on series

    I need something like $\displaystyle \lim \limits_{n \to \infty} a_n=\infty$ and $\displaystyle \lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$
    So take $\displaystyle a_n=n$

    $\displaystyle \frac{a_{n+1}}{a_n}=\frac{n+1}{n}=1+\frac 1n$ which goes to 1 as n goes to infinity.
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