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.