# Series and Sequence Covergence

• Sep 11th 2008, 11:20 AM
johnny0987
Series and Sequence Covergence
Show that if $\sum |a_{n} - a_{n+1}|<\infty$ then the sequence { $a_{n}$} converges, but not conversely.
• Sep 11th 2008, 12:05 PM
Laurent
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.
• Sep 11th 2008, 12:25 PM
bkarpuz
Quote:

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.
• Sep 11th 2008, 02:04 PM
Laurent
Quote:

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

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... (Headbang)

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 $m$.