Past final exam question for an intro to Real Analysis course:

Let $C > 0$, $0<r<1$ and suppose that $\forall n\in \mathbb N, |x_{n+1} - x_n| \leq Cr^n$. Please help me prove that $(x_n)$ is a Cauchy sequence. (We can assume the $\lim\limits_{n\to\infty} r^n=0,$ for $0<r<1$)

So, I know that a Cauchy series must satisfy $|x_{n}-x_m| < \epsilon$ for any $\epsilon>0, \in \mathbb R$ and for all $n,m \gt H(\epsilon) \in \mathbb N$ Note that there can't be any conditions on n and m (I saw somewhere else someone required $m>n$ which you can only do if you're proving its not at Cauchy sequence, right?)

Another way of doing this is showing that it is contractive (and thus a Cauchy series) if there is a constant $a$ such that $|x_{n+1}-x_n| \leq a|x_n-x_{n-1}|$

Clearly I'm supposed to make use of $\lim\limits_{n\to\infty} r^n=0,$ for $0<r<1$... But I don't even know how where to start with this. As I am bumbling through this problem, a more thorough answer would be much appreciated. Thanks!

  • $\begingroup$ Well you solved it. $\endgroup$ – user67773 May 1 '13 at 9:45
  • $\begingroup$ @Uma kant, I haven't really done much actually. I have simply stated some theorems about Cauchy and contractive sequences, which doesn't get me very far. $\endgroup$ – Christian May 1 '13 at 9:47
  • 1
    $\begingroup$ @Umakant No, not yet completely. Neither can we find $a$ with $|x_{n+1}-x_n|\le a|x_n-x_{n-1}|$ for almost all $n$ (we might have $x_n=x_{n-1}$ infinitely often); nor is $\lim_{n\to\infty} Cr^n=0$ sufficient (note that we'd also have $\lim_{n\to\infty}\frac Cn=0$, but $x_n=\ln n$ is not Cauchy). $\endgroup$ – Hagen von Eitzen May 1 '13 at 9:50
  • 1
    $\begingroup$ You can assume $n > m$ without loss of generality since $|x_{n} - x_{m}| = |x_{m} - x_{n}|$. $\endgroup$ – Vincent Pfenninger May 1 '13 at 9:51

Let $\epsilon>0$ be given. Then we can find $N$ such that $r^N\frac C{\color{red}{1-r}}<\frac\epsilon2$ because $r^n\color{red}\to0$ as $n\to \infty$. Then for any $n>N$ we have $$\begin{align} |x_n-x_N|&\le|x_{N+1}-x_N|+\ldots+ |x_n-x_{n-1}|\\&\le Cr^N+Cr^{N+1}+\ldots +Cr^{n-1} \\&=Cr^N\cdot(1+r+\ldots + r^{n-N-1}) \\&\color{red}<Cr^N\sum_{k=0}^\infty r^k \color{red}= Cr^N\cdot\frac1{1-r}<\frac\epsilon2,\end{align}$$ hence for $n,m>N$ $$ |x_n-x_m|\le|x_n-x_N|+|x_m-x_N|<\frac\epsilon2+\frac\epsilon2=\epsilon.$$

Remark: I've marked all places in red where we made use of $0<r<1$.

  • $\begingroup$ +1. a little modification of your proof, it is as good as yours: $|x_n-x_N|\leq Cr^N+Cr^{N+1}+\ldots +Cr^{n-1}=C(r^N+r^{N+1}+\ldots+r^{n-1})=C\frac{r^N(1-r^{n-N})}{1-r}\leq C\frac{r^N}{1-r}<\epsilon/2$. $\endgroup$ – Eric Jan 31 '17 at 6:50
  • $\begingroup$ Also, the "final" step of the proof can also be shown by: for $m,n>N$, without lost of generality, let $m$ be the bigger number(i.e. $m>n$), then $|x_m-x_n|\leq |x_n-x_{n+1}|+\ldots+|x_{m-1}+x_m|\leq|x_N-x_{N+1}|+|x_{N+1}-x_{N+2}|+\ldots+(|x_n-x_{n+1}|+\ldots+|x_{m-1}+x_m|)=|x_m-x_N|<\epsilon/2$. $\endgroup$ – Eric Jan 31 '17 at 7:03
  • $\begingroup$ (in this way, the $N$ need not choose to satisfy $C\frac{r^N}{1-r}<\epsilon/2$. Only need to be less than $\epsilon$) $\endgroup$ – Eric Jan 31 '17 at 7:04

Hint: $|x_n-x_m|\le |x_n- x_{n-1}|+|x_{n-1}-x_{n-2}|+\dots+|x_{m+1}-x_{m}|$.


We don't need to assume $r^n\xrightarrow{n\to\infty}0$. Since $|r|<1$, $$\sum_{n=1}^\infty r^n=\frac r{1-r}$$ and hence $r^n\xrightarrow{n\to\infty}0$.

The desired claim can easily be generalized: Let $(E,d)$ be a metric space, $(x_n)_{n\in\mathbb N}\subseteq E$ and $(\varepsilon_n)_{n\in\mathbb N}\subseteq\mathbb R$ be summable with $$d(x_n,x_{n+1})<\varepsilon_n\;\;\;\text{for all }m,n\ge N\tag1$$ for some $N\in\mathbb N$.

We show that $(x_n)_{n\in\mathbb N}$ is Cauchy: Let $\varepsilon>0$. Since $(\varepsilon_n)_{n\in\mathbb N}$ is summable, $\left(\sum_{i=1}^n\varepsilon_i\right)_{n\in\mathbb N}$ is Cauchy and hence $$\sum_{i=m+1}^{n}\varepsilon_i=\left|\sum_{i=1}^m\varepsilon_i-\sum_{i=1}^n\varepsilon_i\right|<\varepsilon\;\;\;\text{for all }n\ge m\ge N_1\tag2$$ and hence $$d(x_m,x_n)=\sum_{i=m}^{n-1}d(x_i,x_{i+1})<\sum_{i=m}^{n-1}\varepsilon_i<\varepsilon\;\;\;\text{for all }n\ge m>\max(N,N_1).\tag3$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.