Prove or disprove $\sum_{k=0}^\infty\|y_{k+1}-y_k\|<\infty$ if $y_{k+1}=\big(I_n-\frac{ax_kx_k^{\rm T}}{b+\|x_k\|^2}\big)y_k$ 
For all $k\in\mathbb{N}$, let $x_k,\ y_k\in\mathbb{R}^n$, and assume that 
  \begin{equation}
y_{k+1}=\left(I_n-\frac{ax_kx_k^{\rm T}}{b+\|x_k\|^2}\right)y_k,~~~~~~~~~~~~(1)
\end{equation}
  where $a\in(0,1)$, and $b>0$. Prove or disprove by counterexample that 
  $$\sum_{k=0}^\infty\|y_{k+1}-y_k\|<\infty.$$


I can show the summablity for the case where $n=1$: Since $a\in(0,1)$, it follows that
\begin{align}
|y_{k+1}|-|y_k|&=\left|1-\frac{ax_k^2}{b+x_k^2}\right||y_k|-|y_k|\\
&=-\frac{ax_k^2}{b+x_k^2}|y_k|,
\end{align}
which is nonpositive. Thus,
\begin{align}
0\leq \sum_{k=0}^\infty\frac{ax_k^2}{b+x_k^2}|y_k|\leq\sum_{k=0}^\infty\left(|y_{k}|-|y_{k+1}|\right)\leq |y_{0}|-\lim_{k\to\infty}|y_k|\leq |y_{0}|.
\end{align}
Since, in the above expression, the lower and upper bounds exist, it follows that
\begin{equation}
\sum_{k=0}^\infty\frac{ax_k^2}{b+x_k^2}|y_k|<\infty,
\end{equation}
Thus, it follws from (1) that
\begin{equation}
\sum_{k=0}^\infty|y_{k+1}-y_k|=\sum_{k=0}^\infty\frac{ax_k^2}{b+x_k^2}|y_k|<\infty.
\end{equation}

I do not know how to extend this to the case where $n>1$. Any help is appreciated.
 A: Here $$ y_{k+1} = \bigg(I - \frac{a}{b_k +1} x_kx_k^T \bigg) y_k$$
where $|x_k|=1,\ b_k>0,\ 0<a<1$. Here $x_kx_k^T$ is a projection.
So clearly $|y_{k+1}|\leq y_k$ and $L:=\sum_k\ |y_k-y_{k-1}|$ is
length of piecewise
broken curve.
(1) If $x_k \in \{ e_1,\cdots,e_n\}$ where $e_i$
is a canonical
basis, then $L\leq \|y_0\|_1$.
(2) Extreme case : Assume that $a=1,\ b_k=0$ :
If $S_k$ is a radius $\frac{|y_k|}{2}$ sphere whose center is
$\frac{y_k}{2}$, then $y_{k+1}$ is any
point in $S_k$.
If $|y_0|=1$, then assume that $\angle\ (y_i,y_{i+1})=\theta_i$,
then $$ |y_{i+1}|=\cos\ \theta_i |y_i| =\prod_{k=0}^i\ \cos\
\theta_k
$$ and
$$  |y_i-y_{i+1}| =|y_i|\sin\ \theta_i = \sin\ \theta_i\prod_{k=0}^{i-1}\ \cos\
\theta_k $$
Hence $$ L =\sum_i\ \sin\ \theta_i\prod_{k=0}^{i-1}\ \cos\ \theta_k
$$
If $0<C<\theta_i<\frac{\pi}{2}-C$, then $L$ is finite. If $\theta_i\rightarrow 0$, then it depends on $\theta_i$ :
If $\sum_i\ \theta_i <\infty$, then $L$ is finite by the above calculation : $\{y_i\}$ can be viewed as enumeration of points in curve of finite length.
If $\sum_i\ \theta_i=\infty,\ \sum_i\ \theta_i^2<\infty$, then $L$
is infinite : As an example, it has an infinite turn around a fixed circle.
If $  \sum_i\ \theta_i= \sum_i\ \theta_i^2 =\infty$, then $L$ is
finite : $y_k$ goes to origin.
If $\theta_i\rightarrow \frac{\pi}{2}$, then $\cos\ \theta_i\rightarrow 0$ so that $L$
 is finite. 
(3) general case can be covered by (2) :
Proof : Let $Y=(I-x_k' (x_k')^T)X$ and $y_{k+1}= (I-A_k
x_kx_k^T)y_k,\ X=y_k ,\ 0<A_k<1$
Note that there is $x_k'$ s.t. $$ |Y-X |=|y_{k+1}-y_k|
$$ 
 Hence we have two piecewise broken curves : One is the curve in (2), we denote it as $y_k'$, and another is that of $y_k$.
Here $y_k'$ is plotted by $y_k$. If $y_k'$ is the origin, then in next
time, from $y_{k+1}$, we can not plot $y_{k+1}'$. So remaining thing is to show that $|Y|\geq |y_{k+1}|$.
Note that $\Delta oYX $ is a right triangle at $Y$ s.t. $y_{k+1}$ is
interior point of the triangle. 
If we extend a segment $[Yy_{k+1}]$, then it intersect at $Z\in
[oX]$. Since $\Delta XYy_{k+1}$ is isosceles, then $\Delta XYZ$ is
acute triangle. Hence we complete the proof. 
