# Making sure if it is Cauchy

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?

• 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). – Dirk Nov 26 '18 at 5:31
• 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). – Matt A Pelto Nov 26 '18 at 5:44

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.

• 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. – Michael Lee Nov 26 '18 at 0:55
• 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. – Chris Custer Nov 26 '18 at 1:08
• 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. – Michael Lee Nov 26 '18 at 1:08
• Oh yeah. My mistake. – Chris Custer Nov 26 '18 at 1:10

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.

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.

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.

• +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 $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$. – Matt A Pelto Nov 26 '18 at 1:28

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.

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