1. ## Strictly contracting sequences

if $\displaystyle (a_n)$ is strictly contracting then $\displaystyle \sum_{n=1}^\infty |a_{n+1}-a_n|$ 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 $\displaystyle |a_{n+1}-a_n| \leq L|a_n-a_{n-1}|$ but i'm not sure how that really helps.

2. 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 $\displaystyle |a_{n+1}-a_n| \leq L|a_n-a_{n-1}|$ but i'm not sure how that really helps.
And remember that $\displaystyle {\color{red}0<L<1}$

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

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

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

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

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

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

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

Let $\displaystyle a_n=L^n$

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

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

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

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

The statement is true!

Thanks again Moo.

4. Originally Posted by Showcase_22
$\displaystyle |a_2-a_1|\lim_{N \rightarrow \infty}\sum_{n=0}^{N-1}L^n$

I would use The Ratio Test (or L'Hopital's) on this:
Why not just use facts about geometric series?

5. umm, what facts?

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

So how would you do it?

6. 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

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

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

7. 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):

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

I think that's better