Is this a correct proof of the following problem:

Let $\{ x_n \}$ be a sequence such that there exist $A >0$ and $C \in (0,1)$ for which

$$|x_{n+1}-x_n| \leq AC^n$$

for any $n \geq 1$. Show that $\{x_n\}$ is Cauchy.

My proof (hopefully):

  1. The sequence is Cauchy if: $(\forall \epsilon >0)(\exists N)(\forall m,n \geq N)(|x_m -x_n|<\epsilon)$
  2. We will be able to select $N$ for any $\epsilon$ selected provided that $AC^n$ goes to zero (i.e. if $AC^n$ goes to zero as $n$ increases then for any selected $\epsilon$ we are able to select large enough $N$, beyond which the absolute difference between $x_m$ and $x_n$ will be epsilon or less).
  3. We also need monotnic sequence for the above to be true. In this case the sequence is monotonically decreasing. Because the absolute difference between the subsequent terms in the sequence becomes smaller and smaller ($\lim_{n \rightarrow \infty} C^n = 0$).

Thus, due to $2$ and $3$. The sequence is Cauchy.

  • $\begingroup$ I think there are problems with $2$ and $3$. Knowing that $AC^n\to 0$ I don't see why you can conclude that $|x_m- x_n|\to 0$. $\endgroup$
    – Arnaldo
    Mar 28, 2017 at 20:03
  • $\begingroup$ I do not conclude that. Knowing $AC^n \rightarrow 0$ means that the difference between consecutive values in the sequence gets smaller. So knowing that it goes to zero also allows to conclude that for any selection of $\epsilon$ we will be able to select large enough $N$ for which the absolute difference between any two values in the sequence beyond that N will be less than that $\epsilon$ $\endgroup$
    – Naz
    Mar 28, 2017 at 20:06
  • $\begingroup$ Take into account what you just said, you can conclude that $|x_{n+1}-x_n|<\epsilon$ and $|x_{m+1}-x_m|<\epsilon$, but why $|x_m-x_n|<\epsilon$? $\endgroup$
    – Arnaldo
    Mar 28, 2017 at 20:08
  • $\begingroup$ because the sequence is monotonically decreasing. And so beyond $N$ if $m < n$, then since $|x_m - x_{m+1}| < \epsilon$ we will have that $|x_m - x_n| < \epsilon$ because $x_n \leq x_{m+1}$ $\endgroup$
    – Naz
    Mar 28, 2017 at 20:11
  • $\begingroup$ I recommend you write your idea using inequalities (like I just did) and see what you get. $\endgroup$
    – Arnaldo
    Mar 28, 2017 at 20:15

1 Answer 1


$$|x_m-x_n|=|(x_m-x_{m-1})+(x_{n-1}-x_n)+(x_{m-1}-x_{n-1})|\\ |x_m-x_n|\le |x_m-x_{m-1}|+|x_{n-1}-x_n|+|x_{m-1}-x_{n-1}|\\ |x_m-x_n|\le A(C^{m}+C^{n})+|x_{m-1}-x_{n-1}|$$

doing the same idea

$$|x_{m-1}-x_{n-1}|\le A(C^{m-1}+C^{n-1})+|x_{m-2}-x_{n-2}|$$

going on with the same idea we get (supposing that $m\le n$):

$$|x_m-x_n|\le A(C^{m}+C^{m-1}+...+C+C^{n}+C^{n-1}+...+C^{n-m+1})+|x_{0}-x_{n-m}|\quad (1)$$


$$|x_k-x_0|=|x_k-x_{k-1}+x_{k-1}-x_0|\le AC^k+|x_{k-1}-x_0|\quad (2)$$

Using $(2)$ in $(1)$ we get:

$$|x_m-x_n|\le A[(C^{m}+C^{m-1}+...+C)+(C^{n}+C^{n-1}+...+C)]=A\left[\frac{C^{m+1}-1}{C-1}+\frac{C^{n+1}-1}{C-1}\right]$$

Once $C^{m+1}-1$ and $C^{n+1}-1$ goes to zero when $m$ and $n$ goes to infinity then we can choose $m$ and $n$ big enough such that

$$\frac{C^{m+1}-1}{C-1}\le\frac{\epsilon}{2A},\\ \frac{C^{n+1}-1}{C-1}\le\frac{\epsilon}{2A}$$

for any $\epsilon$.


$$|x_m-x_n|\le \epsilon$$


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