Different way of expressing rate of convergence for a sequence Suppose that we have a sequence given by an iterative method: $$x_{n+1} = \phi(x_n)$$
where this sequence converges to a fixed point $\alpha$.
Also, suppose that the sequence has the following rate of convergence $\lambda$:
$$\lim_{n\to \infty} \frac{|x_{n+1} - \alpha|}{|x_{n} - \alpha|^k} = \lambda$$
where $k$ is the order of convergence.

It is asked us to prove that: 
  $$\lim_{n\to \infty} \frac{|x_{n+1} - x_n|}{|x_{n} - x_{n-1}|^k} = \lambda$$

I tried to approach this problem using the triangular inequality and squeezing theorem for sequences. However, it didn't work and it is not clear for me if this approach is useful to solve this problem.
I also tried to suppose that $\phi$ is $k$ times differentiable and I concluded that $\phi^{(l)}(\alpha) = 0$ for $l = 1, \cdots , k-1$ using Taylor expansion. But I don't know how it helps.
I appreciate news ideas and hints to solve this problem. Thanks in advance!
 A: Suppose $k<1$. Then
$$\frac{\left|x_{n+1}-x_n\right|}{\left|x_n-x_{n-1}\right|^k}=\frac{\left|(x_{n+1}-\alpha)-(x_n-\alpha)\right|}{\left|(x_n-\alpha)-(x_{n-1}-\alpha)\right|^k}=\frac{\left|x_n-\alpha\right|}{\left|x_{n-1}-\alpha\right|^k}\frac{\left|\frac{(x_{n+1}-\alpha)}{(x_n-\alpha)^k}(x_n-\alpha)^{k-1}-1\right|}{\left|\frac{(x_n-\alpha)}{(x_{n-1}-\alpha)^k}(x_{n-1}-\alpha)^{k-1}-1\right|^k}$$
And
$$\lim_{n\rightarrow\infty}\left|\frac{(x_{n+1}-\alpha)}{(x_n-\alpha)^k}(x_n-\alpha)^{k-1}-1\right|=1$$
Because $\frac{(x_{n+1}-\alpha)}{(x_n-\alpha)^k}$ is bounded by, say $\pm2\lambda$ and $(x_n-\alpha)^{k-1}\rightarrow0$. Similarly
$$\lim_{n\rightarrow\infty}\left|\frac{(x_n-\alpha)}{(x_{n-1}-\alpha)^k}(x_{n-1}-\alpha)^{k-1}-1\right|=1$$
So
$$\lim_{n\rightarrow\infty}\frac{\left|x_{n+1}-x_n\right|}{\left|x_n-x_{n-1}\right|^k}=\lim_{n\rightarrow\infty}\frac{\left|x_n-\alpha\right|}{\left|x_{n-1}-\alpha\right|^k}=\lambda$$
Now, $k<1$ implies
$$\lim_{n\rightarrow\infty}\frac{\left|x_{n+1}-\alpha\right|}{\left|x_n-\alpha\right|}=\lim_{n\rightarrow\infty}\frac{\lambda}{\left|x_n-\alpha\right|^{1-k}}=\infty$$
So I don't think that's consistent with the hypothesis of the problem. Also if $k=1$ then
$$\lim_{n\rightarrow\infty}\frac{\left|x_{n+1}-\alpha\right|}{\left|x_n-\alpha\right|}=\lambda<1$$
Or we don't get convergence, so
$$\frac{\left|x_{n+1}-x_n\right|}{\left|x_n-x_{n-1}\right|}=\frac{\left|x_n-\alpha\right|}{\left|x_{n-1}-\alpha\right|}\frac{\left|\frac{(x_{n+1}-\alpha)}{(x_n-\alpha)}-1\right|}{\left|\frac{(x_n-\alpha)}{(x_{n-1}-\alpha)}-1\right|}$$
Now, this could be a problem: what if $\frac{(x_{n+1}-\alpha)}{(x_n-\alpha)}=\lambda$ but $\frac{(x_n-\alpha)}{(x_{n-1}-\alpha)}=-\lambda$? Like if
$$x_n=\alpha+\lambda^n\cos\left((2n+1)\frac{\pi}4\right)$$
In this case I think the theorem is busted :(  
So I think the theorem is valid for $k>1$, not meaningful if $k<1$ and have offered a counterexample if $k=1$.
