Series and Sequence Covergence

**johnny0987** · September 11th 2008, 11:20 AM

Show that if $\sum |a_{n} - a_{n+1}|<\infty$ then the sequence { $a_{n}$ } converges, but not conversely.

**Laurent** · September 11th 2008, 12:05 PM

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.

**bkarpuz** · September 11th 2008, 12:25 PM

Originally Posted by johnny0987

Show that if $\sum |a_{n} - a_{n+1}|<\infty$ then the sequence { $a_{n}$ } converges, but not conversely.

Let
$ \sum\limits_{i=1}^{\infty}a(i)<\infty, $
then we know that
$ \lim\limits_{n\to\infty}a(n)=0 $
provided that $\{a(n)\}_{n\in\mathbb{N}}$ is a nonnegative sequence of reals.

From your question, we see that
$ \lim\limits_{n\to\infty}|a(n)-a(n+1)|=0 $
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
$ 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
$ |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
$ \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))| $
............................. $ \geq\sum\limits_{i=1}^{\infty}|a(2i)-a(2i+1)| $
............................. $ =\sum\limits_{i=1}^{\infty}\frac{2}{2i+1}=\infty. $
On the other hand, we have $\lim\nolimits_{n\to\infty}a(n)=0$ .
This completes the proof.

**Laurent** · September 11th 2008, 02:04 PM

Originally Posted by bkarpuz

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 $a_n\to_n+\infty$ ...

**bkarpuz** · September 11th 2008, 08:48 PM

Originally Posted by Laurent

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 $m$ .

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