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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.

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    $\begingroup$ 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: $\left|\frac1m-\frac1n\right|\le\left|\frac1m\right|+\left|\frac1n\right|\le\frac\epsilon2+\frac\epsilon2=\epsilon$. $\endgroup$ Commented Sep 5, 2019 at 1:41

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Holdsworth
    Commented Sep 5, 2019 at 2:04
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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?

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  • $\begingroup$ It's all about the SEQUENCE $(x_n)$, and NOT about any SERIES $\sum_{n\geq 1}x_n$. $\endgroup$ Commented Sep 5, 2019 at 3:46

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