Q:

Column Stochastic Matrix

Let $$A =\begin{pmatrix}0&0 &0.5&0.5&0\\1/3&0&0&0&0\\1/3&1/2&0&1/2&1\\1/3&1/2&0&0&0\\0&0&1/2&0&0\end{pmatrix}$$
and $$S=\dfrac{1}{5}\begin{pmatrix}1&1 &1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\end{pmatrix}.$$
$M=0.85A+0.15S$ is column stochastic and absolute value of second largest eigenvalue of $M \approx 0.6113.$ I have calculated the eigenvector of $M$ corresponding to eigenvalue $1$ to be $q=(0.2371, 0.0972,0.3489, 0.1385,0.1783)^T.$
If $z=(0.2, 0.3, 0.4, 0.05, 0.05)^T,$ it is known that $\text{lim}_{k\to \infty}\dfrac{\lVert M^{k}z-q \rVert_1 }{ \lVert M^{k-1}z-q \rVert_1 }= 0.6113,$ where $\lVert .  \rVert_1 =$ sum of modulus of the entries of a column vector. 
For $k=50,\dfrac{\lVert M^{k}z-q \rVert_1}{ \lVert M^{k-1}z-q \rVert_1}=\dfrac{7.8781\times 10^{-12}}{1.2888\times 10^{-11}} \approx 0.615 $
But for $k=70, \dfrac{\lVert M^{k}z-q \rVert_1 }{ \lVert M^{k-1}z-q \rVert_1}= \dfrac{5.8287\times 10^{-16}}{4.4409\times 10^{-16}} =1.3125$ I have also tested larger values of $k$ and the ratio seems to converge to $1$ instead. 
Could anyone please advise?  Thank you. 

A:

I feel it is the accuracy.  You already got to $10^{-16}$ range in one of your calculations, but your $q$ is likely a double-precision floating number, which is accurate only up to 15 decimal points.