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Math Help - Series and Sequence Covergence

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    Series and Sequence Covergence

    Show that if \sum |a_{n} - a_{n+1}|<\infty then the sequence { a_{n}} converges, but not conversely.
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    Notice that a_n=a_0+\sum_{k=1}^{n-1} (a_{k+1}-a_k), so that (a_n)_{n\geq 0} converges if, and only if \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 \sum_n\frac{(-1)^n}{n}, for instance), you can deduce a sequence (a_n)_n which converges, but such that \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 \sum |a_{n} - a_{n+1}|<\infty then the sequence { a_{n}} converges, but not conversely.
    Let
    <br />
\sum\limits_{i=1}^{\infty}a(i)<\infty,<br />
    then we know that
    <br />
\lim\limits_{n\to\infty}a(n)=0<br />
    provided that \{a(n)\}_{n\in\mathbb{N}} is a nonnegative sequence of reals.

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

    For the converse part, it suffices to give a counter example.
    Let
    <br />
a(n):=<br />
\begin{cases}<br />
\frac{1}{n+1},&n\text{ is even}\\<br />
-\frac{1}{n},&n\text{ is odd}.<br />
\end{cases}<br />
    Then, we see that
    <br />
|a(n)-a(n+1)|=<br />
\begin{cases}<br />
\frac{2}{n+1},&n\text{ is even}\\<br />
\frac{2(n+1)}{n(n+2)},&n\text{ is odd}.<br />
\end{cases}<br />
    And therefore, we get
    <br />
\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))|<br />
    ............................. <br />
\geq\sum\limits_{i=1}^{\infty}|a(2i)-a(2i+1)|<br />
    ............................. <br />
=\sum\limits_{i=1}^{\infty}\frac{2}{2i+1}=\infty.<br />
    On the other hand, we have \lim\nolimits_{n\to\infty}a(n)=0.
    This completes the proof.
    Last edited by bkarpuz; September 12th 2008 at 03:40 AM. Reason: Mistake is highlighted.
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  4. #4
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    Quote Originally Posted by bkarpuz View Post
    There exists \varepsilon>0 and n_{0}\geq1 such that |a(n)-a(n+1)|<\varepsilon for all n\geq n_{0}, this implies that \{a(n)\}_{n\in\mathbb{N}} is a Cauchy sequence (see the definition of Cauchy sequences and let m=n+1), and Cauch sequences are always convergent in complete spaces ( \mathbb{R} is complete).
    Beware! |a(n)-a(n+1)|\to_n 0 does not imply that (a(n))_n is a Cauchy sequence.

    For instance, take a_n=\sqrt{n}. Then 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 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! |a(n)-a(n+1)|\to_n 0 does not imply that (a(n))_n is a Cauchy sequence.

    For instance, take a_n=\sqrt{n}. Then 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 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 m=n+1, however, for a Cauchy sequence we must have the same conclusion for all m,n\geq n_{0}, not only for some values of m.
    Last edited by bkarpuz; September 11th 2008 at 09:32 PM. Reason: Remark added.
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