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Thread: Series and Sequence Covergence

  1. #1
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    Series and Sequence Covergence

    Show that if $\displaystyle \sum |a_{n} - a_{n+1}|<\infty$ then the sequence {$\displaystyle a_{n}$} converges, but not conversely.
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    Notice that $\displaystyle a_n=a_0+\sum_{k=1}^{n-1} (a_{k+1}-a_k)$, so that $\displaystyle (a_n)_{n\geq 0}$ converges if, and only if $\displaystyle \sum_k (a_{k+1}-a_k)$ is a convergent series.

    If a series is absolutely convergent, then it converges. However the converse is not true. Using a counterexample of this property (like $\displaystyle \sum_n\frac{(-1)^n}{n}$, for instance), you can deduce a sequence $\displaystyle (a_n)_n$ which converges, but such that $\displaystyle \sum_n |a_{n+1}-a_n|$ diverges.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Talking

    Quote Originally Posted by johnny0987 View Post
    Show that if $\displaystyle \sum |a_{n} - a_{n+1}|<\infty$ then the sequence {$\displaystyle a_{n}$} converges, but not conversely.
    Let
    $\displaystyle
    \sum\limits_{i=1}^{\infty}a(i)<\infty,
    $
    then we know that
    $\displaystyle
    \lim\limits_{n\to\infty}a(n)=0
    $
    provided that $\displaystyle \{a(n)\}_{n\in\mathbb{N}}$ is a nonnegative sequence of reals.

    From your question, we see that
    $\displaystyle
    \lim\limits_{n\to\infty}|a(n)-a(n+1)|=0
    $
    holds, i.e. there exists $\displaystyle \varepsilon>0$ and $\displaystyle n_{0}\geq1$ such that $\displaystyle |a(n)-a(n+1)|<\varepsilon$ for all $\displaystyle n\geq n_{0}$, this implies that $\displaystyle \{a(n)\}_{n\in\mathbb{N}}$ is a Cauchy sequence (see the definition of Cauchy sequences and let $\displaystyle {\color{magenta}{m=n+1}}$), and Cauch sequences are always convergent in complete spaces ($\displaystyle \mathbb{R}$ is complete).

    For the converse part, it suffices to give a counter example.
    Let
    $\displaystyle
    a(n):=
    \begin{cases}
    \frac{1}{n+1},&n\text{ is even}\\
    -\frac{1}{n},&n\text{ is odd}.
    \end{cases}
    $
    Then, we see that
    $\displaystyle
    |a(n)-a(n+1)|=
    \begin{cases}
    \frac{2}{n+1},&n\text{ is even}\\
    \frac{2(n+1)}{n(n+2)},&n\text{ is odd}.
    \end{cases}
    $
    And therefore, we get
    $\displaystyle
    \sum\limits_{i=1}^{\infty}|a(i)-a(i+1)|=\sum\limits_{i=1}^{\infty}|a(2i)-a(2i+1)|+\sum\limits_{i=1}^{\infty}|a(2i+1)-a(2(i+1))|
    $
    .............................$\displaystyle
    \geq\sum\limits_{i=1}^{\infty}|a(2i)-a(2i+1)|
    $
    .............................$\displaystyle
    =\sum\limits_{i=1}^{\infty}\frac{2}{2i+1}=\infty.
    $
    On the other hand, we have $\displaystyle \lim\nolimits_{n\to\infty}a(n)=0$.
    This completes the proof.
    Last edited by bkarpuz; Sep 12th 2008 at 03:40 AM. Reason: Mistake is highlighted.
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  4. #4
    MHF Contributor

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    Quote Originally Posted by bkarpuz View Post
    There exists $\displaystyle \varepsilon>0$ and $\displaystyle n_{0}\geq1$ such that $\displaystyle |a(n)-a(n+1)|<\varepsilon$ for all $\displaystyle n\geq n_{0}$, this implies that $\displaystyle \{a(n)\}_{n\in\mathbb{N}}$ is a Cauchy sequence (see the definition of Cauchy sequences and let $\displaystyle m=n+1$), and Cauch sequences are always convergent in complete spaces ($\displaystyle \mathbb{R}$ is complete).
    Beware! $\displaystyle |a(n)-a(n+1)|\to_n 0$ does not imply that $\displaystyle (a(n))_n$ is a Cauchy sequence.

    For instance, take $\displaystyle a_n=\sqrt{n}$. Then $\displaystyle a_{n+1}-a_n=\sqrt{n+1}-\sqrt{n}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{ n}}\to_n 0$, whereas $\displaystyle a_n\to_n+\infty$...
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  5. #5
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Laurent View Post
    Beware! $\displaystyle |a(n)-a(n+1)|\to_n 0$ does not imply that $\displaystyle (a(n))_n$ is a Cauchy sequence.

    For instance, take $\displaystyle a_n=\sqrt{n}$. Then $\displaystyle a_{n+1}-a_n=\sqrt{n+1}-\sqrt{n}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{ n}}\to_n 0$, whereas $\displaystyle a_n\to_n+\infty$...
    Yes, you are completely right, I missed it. :S
    It could be an increasing concave sequence...

    Remark. In my previous message, my mistake is fixing $\displaystyle m=n+1$, however, for a Cauchy sequence we must have the same conclusion for all $\displaystyle m,n\geq n_{0}$, not only for some values of $\displaystyle m$.
    Last edited by bkarpuz; Sep 11th 2008 at 09:32 PM. Reason: Remark added.
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