# Proof verification - Cauchy sequence

Consider the sequence $$(x_n)$$ in $$\mathbb R$$ with the following property:$$\forall n \in \mathbb N_0: |x_{n+1}-x_n | \leq \frac{1}{n(n+1)}$$ Prove that the sequence $$x_n$$ is convergent.

I have already proven the fact that: $$\forall m \in \mathbb N_0:\forall n \in \mathbb N_0: |x_m - x_n| \leq \bigg{\lvert}\frac{1}{m}-\frac{1}{n}\bigg{\rvert}$$ So assume this true. Now let $$\epsilon > 0$$ and $$n_0 > \frac{1}{\epsilon}$$ with $$n_0 \in \mathbb N$$. Also let $$m \ge n_0$$ and $$n \ge n_0$$ with both $$m,n \in \mathbb N$$. There are two cases: $$n \ge m$$ and $$m \ge n$$.

1) $$n \ge m$$:

Because $$m \ge n_0$$ and $$n_0 > \frac{1}{\epsilon}$$, it follows that $$m > \frac{1}{\epsilon}$$ and thus $$\frac{1}{m} < \epsilon$$. Because $$\frac{1}{n} > 0$$, it follows that $$\frac{1}{m}-\frac{1}{n} < \epsilon$$. Since $$n \ge m$$ and consequently $$\frac{1}{m} -\frac{1}{n} \ge 0$$, I can take the absolute value without problem: $$\bigg{\lvert}\frac{1}{m}-\frac{1}{n}\bigg{\rvert} < \epsilon$$. Using the above proven statement it follows that $$|x_m-x_n| < \epsilon$$.

2) $$m \ge n$$

Because $$n \ge n_0$$ and $$n_0 > \frac{1}{\epsilon}$$, it follows that $$n > \frac{1}{\epsilon}$$ and thus $$\frac{1}{n} < \epsilon$$. Because $$\frac{1}{m} > 0$$, it follows that $$\frac{1}{n}-\frac{1}{m} < \epsilon$$. Since $$m \ge n$$ and consequently $$\frac{1}{n} -\frac{1}{m} \ge 0$$, I can take the absolute value without problem: $$\bigg{\lvert}\frac{1}{n}-\frac{1}{m}\bigg{\rvert} < \epsilon$$. Since $$\bigg{\lvert}\frac{1}{n}-\frac{1}{m}\bigg{\rvert} = \bigg{\lvert}\frac{1}{m}-\frac{1}{n}\bigg{\rvert}$$, I can use the above proven statement again so that $$|x_m-x_n| < \epsilon$$.

I have therefore proven in both cases that for every $$\epsilon > 0$$ there exists an $$n_0 \in \mathbb N$$ so that for all $$n \ge n_0$$ and for all $$m \ge n_0$$: $$|x_m - x_n| < \epsilon$$. Thus, the sequence $$(x_n)$$ is a Cauchy sequence by definition and therefore also convergent. $$\square$$

Is this a valid proof ? I get the feeling that I used an illegitimate case seperation because of those confusing indices.

• There is never anything wrong with splitting a problem into multiple cases provided you handle all cases, but there is no need to explicit write out both cases since this is symmetric in $m$ and $n$. And you can cover both cases by letting $m,n\ge\frac2\epsilon$, which gives by triangle inequality: $\left|\frac1m-\frac1n\right|\le\left|\frac1m\right|+\left|\frac1n\right|\le\frac\epsilon2+\frac\epsilon2=\epsilon$. – Simply Beautiful Art Sep 5 '19 at 1:41

I will start by stating what @Simply Beautiful Art said and add a bit more into it.

It's common to state the argument for doing such an act(as assuming n>=m), which would be without loss of generality and not handling the other case in which m>=n

and when referring to I can take out the absolute value without problem - I'd personally write, by absolute value properties, a>=0 <=> |a|=a, it's explaining why could you do such an act, of course I'd point that this a I am referring to - is really non-negative.

Setting this aside, truly the triangle inequality |a-b|<=|a|+|b| does shorten the proof here substantially and is often used, also for example when proving that every sequence that is Cauchy sequence is equivalent to the formal definition of a convergent sequence.

• It's important to understand WHEN NOT TO USE without loss of generality, in one word when I wouldn't use it - in asymmetric cases. – Holdsworth Sep 5 '19 at 2:04

I am wondering about this proof. Suppose we have a sequence $$x_n = \frac{1}{n}$$, which make the property above hold, i.e. $$\forall n \in \mathbb{N}_0 : |x_{n+1} - x_n| \le \frac{1}{n(n+1)}$$

But this sequence $$x_n = \frac{1}{n}$$, i.e., the reciprocals of the positive integers, produces a divergent series. Does it mean the proof is a pseudo proposition?

• It's all about the SEQUENCE $(x_n)$, and NOT about any SERIES $\sum_{n\geq 1}x_n$. – trancelocation Sep 5 '19 at 3:46