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If $|c|<1$, show $\lim\limits_{n\rightarrow\infty}c^n=0.$ Use that fact that $|c^n|\leq\frac{1}{1+nd}\leq\frac{1}{nd}$ where $d>0$ and every index $n$.


Suppose that $|c|<1$ and $|c^n|\leq\frac{1}{1+nd}\leq\frac{1}{nd}$ where $d>0$ and every index $n$. Let $\epsilon>0$. Then apply the Archimedean Property, there exists an index $N$ such that for all $n\geq N$. Then let $\epsilon'=d\epsilon$. Then we have $$\lim\limits_{n\rightarrow\infty}|c^n-0|\leq\lim\limits_{n\rightarrow\infty}\left|\frac{1}{nd}-0\right|\leq\frac{1}{nd}\leq\frac{1}{Nd}<\epsilon'/d=\epsilon$$ This shows that $\lim\limits_{n\rightarrow\infty}c^n=0$


Can anyone check my proof? I am not sure cause I just follow the idea from the comparison lemma. Thanks in advanced.

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    $\begingroup$ Where do you get the first inequality? $\endgroup$ – Lutz Lehmann Oct 18 '15 at 21:44
  • $\begingroup$ @LutzL from the question statement $\endgroup$ – Simple Oct 18 '15 at 21:46
  • $\begingroup$ Ok, if it was meant as a given fact and not as a hint for the direction of the proof, then what you did would be correct if you also write down what the relation of $N$ and $ϵ′$ is. $\endgroup$ – Lutz Lehmann Oct 18 '15 at 21:51
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Then apply the Archimedean Property, there exists an index $N$ such that for all $n\geq N$. Then let $\epsilon'=d\epsilon$.

This is not so clear what you are referring to. You did not say how you find $N$, and how this $N$ is related to your $\epsilon$. Putting this aside, we first take a look at your inequality:

$$\lim\limits_{n\rightarrow\infty}|c^n-0|\leq\lim\limits_{n\rightarrow\infty}\left|\frac{1}{nd}-0\right|\leq\frac{1}{nd}\leq\frac{1}{Nd}<\epsilon'/d=\epsilon$$

Note that the first two terms has the limit $\lim_{n\to \infty}$. This mean that these two term is independent of $n$ (You have taken the limit). However, you don't even know if the limit exist (which is what you what to prove), so writing the limit is not correct. But actually you don't need that, just write

$$|c^n-0|\leq \left|\frac{1}{nd}-0\right|\leq\frac{1}{nd}\leq\frac{1}{Nd}<\epsilon'/d=\epsilon . $$

The first two inequality is clear and the third you used $n \ge N$. Then now it is unclear why

$$\frac{1}{Nd}<\frac{\epsilon'}{d}$$

since you haven't relate $N$ and $\epsilon'$. But this inequality is the same as $ N > \frac{1}{\epsilon ' }$. So should write before this inequality: By the Archimedean property, there is $N \in \mathbb N$ so that $ N > \frac{1}{\epsilon'}$, now let $n \ge N$....

Then you get $|c^n - 0| <\epsilon$ whenever $n\ge N$ and so you can conclude.

Remark Let $|c| <1$ and $c\neq 0$. Then $\frac{1}{|c|} >1$. So we write $\frac{1}{|c|} = 1+ d$ for some $d>0$. Then $$\left(\frac{1}{|c|}\right)^n = (1+ d)^n \ge 1+ nd$$ by the Bernoulli's inequality. Thus $$|c|^n \le \frac{1}{1+nd}.$$

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  • $\begingroup$ Thanks. I don't quite follow how to get $1/|c|=1+d$ on the Remark $\endgroup$ – Simple Oct 18 '15 at 22:09
  • $\begingroup$ @Simple :as $\frac{1}{|c|}>1$, I can write $\frac{1}{|c|} =1 + (1/|c| -1)$ and call $d = 1/|c| -1$, which is a positive number. $\endgroup$ – user99914 Oct 18 '15 at 22:12

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