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Context: I done this problem awhile back and was looking through my notes on it and my answer seems incorrect.

Let $(x_n)_{n{\in}\mathbb{N}}$ be a sequence such that $|x_n-x_{n+1}|\;{\le}\;2^{-n}$ holds for every $n\,{\in}\,\mathbb{N}$. Show that $(x_n)_{n{\in}\mathbb{N}}$ is a Cauchy sequence.

My initial answer:

$$|x_n-x_{n+1}|\;{\le}\;2^{-n}{\implies}(x_n)_{n{\in}\mathbb{N}}\;\text{is monotone and}\;2^{-n}\rightarrow0{\implies}(x_n)_{n{\in}\mathbb{N}}\;\text{is bounded. Thus}\;(x_n)_{n{\in}\mathbb{N}}\;\text{is convergent which implies it is a Cauchy sequence}$$

My issue with this answer is I believe "$|x_n-x_{n+1}|\;{\le}\;2^{-n}{\implies}(x_n)_{n{\in}\mathbb{N}}\;\text{is monotone}$" is false. The inequality is not sufficiently strong to impose that the sequence is either increasing or decreasing?

My thoughts are that removing the modulus sign (purely done because I could not get anywhere otherwise) would make it a more valid answer (obviously it would no longer answer the initial question and I still doubt it is entirely correct even with no modulus sign). The reason I doubt that it is correct without the modulus sign is the Wikipedia page on Cauchy sequences briefly talks about the sequence of square roots of natural numbers not being a Cauchy sequence despite the consecutive terms becoming arbitrarily close and I noticed that my argument would suggest that it is a Cauchy sequence.

After this I realised what the wiki was saying and observed that $$\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}} \;\text{diverges since}\sum_{n=1}^{\infty} \frac{1}{n}\;\text{diverges and}\,0\le\frac{1}{n}\le\frac{1}{2\sqrt{n}}\;\text{for}\; n\ge4$$ (the above summation comes from the wikipedia page)

but $$\sum_{n=1}^{\infty} 2^{-n}=1$$

However now I am just lost (probably due to lack of understanding about Cauchy sequences). Doesn't the fact the series converges to 1 mean that you could use the same argument from the Wikipedia page on Cauchy sequences? I actually doubt this is the case somewhat but can't seem to wrap my head around the logic.

Any help would be greatly appreciated. My main issue comes with the modulus sign and also the fact that the series of distances between consecutive terms converge to 1 a (not even sure this is relevant however).

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Yes, you're argument was wrong, since as you said the condition on the absolute value cannot tell you anything about the monotonicity. However you can note that for any $m>0$ you have $$|x_{n+m}-x_n| \leq \sum_{k=n}^{n+m-1}|x_{k+1}-x_k|\leq\sum_{k=n}^{n+m-1} 2^{-k} = 2^{-n}\sum_{k=0}^{m-1}2^{-k}\,.$$ From here you can see that indeed the argument $\sum_{k=1}^\infty 2^{-k}=1$ can be used to prove that the sequence is Cauchy. Indeed you have that $\sum_{k=0}^{m-1}2^{-k}\leq \sum_{k=0}^{\infty}2^{-k}=2$ for all $m\geq 1$. So you get $$|x_{n+m}-x_n|\leq 2\times 2^{-n}$$ which tends to $0$ for $n\to\infty$. This proves that the sequence is Cauchy.

Note that in general, a sufficient condition for a sequence to be Cauchy is that, given that $|x_{n+1}-x_n|\leq \epsilon_n$, $\sum_{n=0}^\infty\epsilon_n \leq \infty$. Indeed you will have $$|x_{n+m}-x_n|\leq \sum_{k=n}^\infty \epsilon_k$$ and the RHS must vanishes for $n\to \infty$.

However such a condition is in general not necessary. For instance the sequence $x_n=(-1)^n\frac{1}{n}$ clearly converges (and so is Cauchy) but $\sum_{n=0}^\infty |x_{n+1}-x_n| = \infty$. It is the case that the condition on the sum is necessary if $x_n$ is a monotonous sequence.

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    $\begingroup$ This is very helpful thanks so much. There is only one part I am struggling to see as of reading it which is on the top line where you shift the indexes of the summation, however I am sure I will wrap my head around it soon enough. Thanks again (I am going to mark this as the answer as soon as it lets me). $\endgroup$
    – user732461
    Jul 30 '20 at 9:33
  • $\begingroup$ I'm just saying $\sum_{k=n}^{n+m-1}2^{-k} = \sum_{k'=0}^{m-1}2^{-n-k'}$ defining $k'=k-n$. Then you can take $2^{-n}$ outside of the sum and rewrite $k$ instead of $k'$. $\endgroup$
    – ECL
    Jul 30 '20 at 9:36

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