Series and Sequence Covergence

• Sep 11th 2008, 11:20 AM
johnny0987
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.
• Sep 11th 2008, 12:05 PM
Laurent
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.
• Sep 11th 2008, 12:25 PM
bkarpuz
Quote:

Originally Posted by johnny0987
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.
• Sep 11th 2008, 02:04 PM
Laurent
Quote:

Originally Posted by bkarpuz
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 $\displaystyle a_n\to_n+\infty$...
• Sep 11th 2008, 08:48 PM
bkarpuz
Quote:

Originally Posted by Laurent
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... (Headbang)

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 $\displaystyle m$.