Strictly contracting sequences

**Showcase_22** · December 17th 2008, 04:01 AM

if $(a_n)$ is strictly contracting then $\sum_{n=1}^\infty |a_{n+1}-a_n|$ converges.

Is this statement true or false?

I decided that this statement was true but I couldn't find a concrete way of justifying it.

I tried using $|a_{n+1}-a_n| \leq L|a_n-a_{n-1}|$ but i'm not sure how that really helps.

**Moo** · December 17th 2008, 04:19 AM

Hello !

Originally Posted by Showcase_22

I decided that this statement was true but I couldn't find a concrete way of justifying it.

I tried using $|a_{n+1}-a_n| \leq L|a_n-a_{n-1}|$ but i'm not sure how that really helps.

And remember that ${\color{red}0<L<1}$

Here is the general idea :
$\sum_{n=1}^\infty |a_{n+1}-a_n|=\lim_{N \to \infty} S_N$ , where $S_N=\sum_{n=1}^N |a_{n+1}-a_n|$

It can easily be shown (by induction) that $|a_{n+1}-a_n| \leq L^{n-1} |a_2-a_1|$

Hence $\sum_{n=1}^N |a_{n+1}-a_n| \leq \sum_{n=1}^N L^{n-1} |a_2-a_1|=|a_2-a_1| \sum_{n=1}^N L^{n-1}=|a_2-a_1| \sum_{n=0}^{N-1} L^n$

Therefore $\sum_{n=1}^\infty |a_{n+1}-a_n|=\lim_{N \to \infty} |a_2-a_1| \sum_{n=0}^{N-1} L^n=|a_2-a_1| \lim_{N \to \infty} \sum_{n=0}^{N-1} L^n$

Is it a converging series ?

Note : I introduced N, because since we don't know if it converges or not, we're normally not allowed to write the infinity in the sum

**Showcase_22** · December 17th 2008, 11:49 PM

$|a_2-a_1|\lim_{N \rightarrow \infty}\sum_{n=0}^{N-1}L^n$

I would use The Ratio Test (or L'Hopital's) on this:

Let $a_n=L^n$

$\frac{|a_{n+1}|}{|a_n|}=\frac{L^{n+1}}{L^n}$

$\frac{|a_{n+1}|}{|a_n|}=\frac{L^n L}{L^n}$

$\frac{|a_{n+1}|}{|a_n|}=L$

Since L is between 0 and 1 ( $0 \leq L \leq 1$ ) then the series is convergent.

The statement is true!

Thanks again Moo.

**ThePerfectHacker** · December 18th 2008, 09:09 AM

Originally Posted by Showcase_22

$|a_2-a_1|\lim_{N \rightarrow \infty}\sum_{n=0}^{N-1}L^n$

I would use The Ratio Test (or L'Hopital's) on this:

Why not just use facts about geometric series?

**Showcase_22** · December 22nd 2008, 07:31 AM

umm, what facts?

I know that I would be adding increasingly smaller terms each time, but the same goes for $\sum_{k=1}^{n} \frac{1}{k}$ and that diverges.

So how would you do it?

**Ziaris** · December 22nd 2008, 09:23 AM

The harmonic series is not a geometric series. It's easy to confuse zeta-type sums with geometric series since they are superficially similar, but note the difference between

$\lim_{A\to\infty}\sum_{n=1}^{A}n^{-s}=1+2^{-s}+3^{-s}+4^{-s}+\cdots$ and $\lim_{A\to\infty}\sum_{n=0}^{A}{k^n}=1+k+k^2+k^3+\ cdots$

Note how in the first sum the number is changing and in the second the number $k$ is constant while the power is changing. Both series in this case are convergent if $s>1$ for the left, and for the geometric series on the right $|k|<1, k\neq 0.$ Since this is the case, you can conclude that given your work that your series is indeed convergent if $0 < L < 1.$ (Note that the harmonic series with a finite upper bound is convergent, so that the series you posted $\sum_{k=1}^{n}\frac{1}{k}$ does not $\to\infty$ if $n$ does not $\to\infty.$ )

**Showcase_22** · December 23rd 2008, 06:36 AM

I knew that the harmonic series wasn't geometric. I was using it as an example (admittedly a bad one! =S) to try and show that a series can tend to infinity even though successive terms get smaller.

I also apologise for miswriting my sum (I was on a library computer and had to write quickly):

$\sum_{k=1}^{\infty} \frac{1}{k} \rightarrow \infty$

I think that's better

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