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In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|\lt {a^n}$ for all natural numbers $n$ and for all positive number $a\lt 1$ then $(x_n)$ is a Cauchy sequence.

This was solved successfully but the question is if $|x_{n+1} - x_n|\lt \frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?

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  • $\begingroup$ Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large). $\endgroup$ – Dirk Nov 26 '18 at 5:31
  • $\begingroup$ Seems to be some sort of error. The proper correction should be either changing "$\frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series). $\endgroup$ – Matt A Pelto Nov 26 '18 at 5:44
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Consider the harmonic series: $\sum_{n=1}^{\infty}\frac 1n$.

$\mid a_{n+1}-a_n\mid=\frac1{n+1}\lt\frac1n$.

But it diverges.

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  • $\begingroup$ This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{n\in \mathbb{N}}$ is indeed Cauchy. $\endgroup$ – Michael Lee Nov 26 '18 at 0:55
  • $\begingroup$ Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $\mid a_{n+1}-a_n\mid$ can be less than $\frac1n$, but the sequence can still not be Cauchy. $\endgroup$ – Chris Custer Nov 26 '18 at 1:08
  • $\begingroup$ The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th. $\endgroup$ – Michael Lee Nov 26 '18 at 1:08
  • $\begingroup$ Oh yeah. My mistake. $\endgroup$ – Chris Custer Nov 26 '18 at 1:10
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No, $\lvert x_{n+1}-x_n\rvert < 1/n$ does not imply that $(x_n)_{n\in \mathbb{N}}$ is Cauchy. Consider $x_n = \sum_{k=1}^n 1/2k$, which does not converge.

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Take $x_n=1+\frac 1 2+\cdots+\frac 1 n$. This is not Cauchy because the harmonic series $1+\frac 1 2+\cdots$ is divergent.

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This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $\sum_{k = 0}^\infty a_k < \infty$ because, if this condition holds,

\begin{align} |x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - \cdots + x_{n + m - 1} - x_{n + m}| \\ &\le |x_{n} - x_{n + 1}| + \cdots + |x_{n + m - 1} - x_{n + m}| \\ &\le a_n + a_{n + 1} + \dots + a_{n + m - 1} \\ &\le \sum_{k = n}^\infty a_k \end{align}

Now convergence of $\sum a_k$ to $A$ means that for any $\varepsilon > 0$ there exists $N$ such that for all $n \ge N$,

$$ \left| A - \sum_{k = 0}^{n - 1} a_k \right| = \sum_{k = n}^\infty a_k < \varepsilon $$

Comparing this with the above, we have for every $n \ge N$ and every $m \ge 0$,

$$ |x_n - x_{n + m}| < \varepsilon $$

Which means the sequence $(x_n)$ is Cauchy.

If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.

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    $\begingroup$ +1 for the most informative response and using \varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$. $\endgroup$ – Matt A Pelto Nov 26 '18 at 1:28
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For each $n \in \mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+\cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=\frac{1}{n+1}$ but the sequence $\{x_n\}_{n=1}^\infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $\mathbb{R}$ is complete.

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This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).

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