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I am learning Metric Fixed Point Theory by Mohammed A Khamsi and William A Kirk.

I need help in understanding a step in the proof of the following theorem(Chapter 3, Theorem 3.2, Page No. 43):

Let $(M,d)$ be a complete metric space, let $T:M\rightarrow M$ be a contraction mapping with Lipschitz constant $k\in(0,1)$, and suppose $x_0\in M$ is the fixed point of $T$. Let $\{\epsilon_n\}$ be a sequence of positive numbers for which $\lim_{n\rightarrow\infty}\epsilon_n=0$, let $y_0\in M$, and suppose $\{y_n\}\subset M$ satisfies $d(y_{n+1},T(y_n))\leq \epsilon_{n}$. Then $\lim_{n\rightarrow\infty}=x_{0}$

In proof we assume $y_0=x$. By applying the triangle inequality, contraction and the assumption $d(y_{n+1},T(y_n))\leq \epsilon_{n}$ we get, $$d(T^{m+1}(x),y_{m+1})\leq\sum_{i=0}^mk^{m-i}\epsilon_{i}$$ Again by applying triangle inequality, we get, $$d(y_{m+1},x_{0})\leq\sum_{i=0}^mk^{m-i}\epsilon_{i}+d(T^{m+1}(x),x_{0})$$ Since $\{\epsilon_n\}\rightarrow 0$, for given $\epsilon> 0$, there exists $N\in\mathbb{N}$ such that for $m\geq N$, $\epsilon_m\leq\epsilon$ $$\sum_{i=0}^mk^{m-i}\epsilon_{i}\leq k^{m-N}\sum_{i=0}^Nk^{N-i}\epsilon_{i}+\epsilon\sum_{i=N+1}^mk^{m-i}\epsilon_{i}$$

I understand up to this step. From this how do I get, $$\lim_{m\rightarrow\infty}\sum_{i=0}^mk^{m-i}\epsilon_{i}\leq\epsilon\left(\frac{k^N-1}{1-k}\right)$$

I am not sure how to make $m$ larger and obtain the limit of the sum. I have trouble in figuring out $k^{m-N}$ and $k^{m-i}$ as $m\rightarrow\infty$. Kindly help. Sorry, if I have not posed my question properly.

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  • $\begingroup$ Got it. Thank you. It is actually k^N not k^(N) - 1. It is a mistake. $\endgroup$ – Vicky Apr 19 '15 at 13:53
  • $\begingroup$ you might wanna edit your post, especially the statement of the theorem. $\endgroup$ – Saaqib Mahmood May 17 '16 at 5:52
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The $\epsilon_i$ at the very end of the displayed inequality

$$\sum_{i=0}^mk^{m-i}\epsilon_{i}\leq k^{m-N}\sum_{i=0}^Nk^{N-i}\epsilon_{i}+\epsilon\sum_{i=N+1}^mk^{m-i}\epsilon_{i}$$

should not be there: it’s

$$\sum_{i=0}^mk^{m-i}\epsilon_{i}\leq k^{m-N}\sum_{i=0}^Nk^{N-i}\epsilon_{i}+\epsilon\sum_{i=N+1}^mk^{m-i}\;.$$

Also, $\epsilon\left(\frac{k^N-1}{1-k}\right)$ can’t be right: $k\in(0,1)$, so the fraction is negative. Here’s a correct calculation:

$$\begin{align*} \lim_{m\to\infty}\sum_{i=0}^mk^{m-i}\epsilon_{i}&\le \lim_{m\to\infty}\left(k^{m-N}\sum_{i=0}^Nk^{N-i}\epsilon_{i}+\epsilon\sum_{i=N+1}^mk^{m-i}\right)\\ &=\lim_{m\to\infty}k^{m-N}\sum_{i=0}^Nk^{N-i}\epsilon_{i}+\epsilon\lim_{m\to\infty}\sum_{i=N+1}^mk^{m-i}\\ &\overset{(*)}=\left(k^{-N}\sum_{i=0}^Nk^{N-i}\epsilon_{i}\right)\lim_{m\to\infty}k^m+\epsilon\lim_{m\to\infty}\sum_{\ell=0}^{m-N-1}k^\ell\\ &=0+\epsilon\sum_{\ell=0}^\infty k^\ell\\ &=\epsilon\left(\frac1{1-k}\right)\;. \end{align*}$$

In the step marked $(*)$ I simply let $\ell=m-i$: $N+1\le i\le m$ is then precisely equivalent to $0\le\ell\le m-N-1$. And of course $m-N-1\to\infty$ as $m\to\infty$, so the last limit on that line is just the sum of the infinite geometric series.

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