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Math Help - Ratio test for series divergence help.

  1. #1
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    Ratio test for series divergence help.

    This is from so called ratio test, which says

    The series Ʃa(n) diverges if abs{a(n+1)/a(n)} ≥ 1 for n≥N where N is some fixed integer.

    I was wondering if I could replace the condition with lim inf abs{a(n+1)/a(n)} ≥ 1.

    So basically what I'm asking is whether these two below are equivalent or not.
    1. abs{a(n+1)/a(n)} ≥ 1 for n≥N where N is some fixed integer
    2. lim inf abs{a(n+1)/a(n)} ≥ 1

    *a(n) means nth term of the sequence, and abs means absolute value.

    Thank you in advance.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Ratio test for series divergence help.

    I think You couldn't. Supposing that is...

    \lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}}|=1 (1)

    ... the behaviour of |\frac{a_{n+1}}{a_{n}}}| can be 'oscillating' around 1 and in this case it isn't |\frac{a_{n+1}}{a_{n}}}| \ge 1 'for n \ge N where N is some fixed integer'...



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  3. #3
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    Re: Ratio test for series divergence help.

    Quote Originally Posted by chsoviz0716 View Post
    This is from so called ratio test, which says

    The series Ʃa(n) diverges if abs{a(n+1)/a(n)} ≥ 1 for n≥N where N is some fixed integer.

    I was wondering if I could replace the condition with lim inf abs{a(n+1)/a(n)} ≥ 1.

    So basically what I'm asking is whether these two below are equivalent or not.
    1. abs{a(n+1)/a(n)} ≥ 1 for n≥N where N is some fixed integer
    2. lim inf abs{a(n+1)/a(n)} ≥ 1

    *a(n) means nth term of the sequence, and abs means absolute value.

    Thank you in advance.
    Actually, the series is only convergent where \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1 \end{align*} and divergent where \displaystyle \begin{align*}  \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| > 1\end{align*}.

    The ratio test is INCONCLUSIVE if \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|  = 1 \end{align*}.
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    MHF Contributor chisigma's Avatar
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    Re: Ratio test for series divergence help.

    The series diverges if \lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}|=1 but is also \forall{n>N}\ |\frac{a_{n+1}}{a_{n}}|\ge 1...



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    Re: Ratio test for series divergence help.

    Quote Originally Posted by chisigma View Post
    I think You couldn't. Supposing that is...

    \lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}}|=1 (1)

    ... the behaviour of |\frac{a_{n+1}}{a_{n}}}| can be 'oscillating' around 1 and in this case it isn't |\frac{a_{n+1}}{a_{n}}}| \ge 1 'for n \ge N where N is some fixed integer'...

    Marry Christmas from Serbia

    \chi \sigma
    Thank you for the reply first.
    The example you took shows even if 2 is satisfied, 1 need not be satisfied, so clearly they are not equivalent.
    What about the other way around?
    Can I conclude that if 1 is satisfied, 2 is satisfied?
    or is there an example which satisfies 1 but doesn't satisfy 2?
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Ratio test for series divergence help.

    Can I conclude that if 1 is satisfied, 2 is satisfied?... or is there an example which satisfies 1 but doesn't satisfy 2? ...

    Let's indicate \rho_{n}= \frac{a_{n+1}}{a_{n}} and consider the possibility \rho_{n}=2+(-1)^{n}. Clearly 1) is satisfied and 2) is not because the limit for n tending to infinity of \rho_{n} doesn't exist...



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    Re: Ratio test for series divergence help.

    Quote Originally Posted by chisigma View Post
    Can I conclude that if 1 is satisfied, 2 is satisfied?... or is there an example which satisfies 1 but doesn't satisfy 2? ...

    Let's indicate \rho_{n}= \frac{a_{n+1}}{a_{n}} and consider the possibility \rho_{n}=2+(-1)^{n}. Clearly 1) is satisfied and 2) is not because the limit for n tending to infinity of \rho_{n} doesn't exist...
    Oh may be you misunderstood what I wrote as 'lim inf'. I didn't mean limit as n goes to infinity but meant 'limit inferior' which is defined to be the infimum of all subsequential limits of the original sequence. I attach the link here.
    Limit superior and limit inferior - Wikipedia, the free encyclopedia

    So even in the case of 2+(-1)^n, lim inf abs{a(n+1)/a(n)} ≥ 1 still holds. In fact, it equals one in this case.
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  8. #8
    MHF Contributor chisigma's Avatar
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    Re: Ratio test for series divergence help.

    For the sequence \rho_{n}=\frac{|a_{n+1}|}{|a_{n}|}= 2+(-1)^{n} it doesn't exist \lim_{n \rightarrow \infty} \rho_{n}...



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