While solving a problem, I thought of kind of "Triangle inequality infinite terms". Clearly for finite terms if $a=x_0$ and $x_n=b$ we have $$|a-b|\leq |a-x_1+x_1-x_2+\cdots+x_{n-1}-b|\leq \sum_{i=1}^n|x_{i-1}-x_{i}| $$ It seems like, under the conditions $\lim_{n\to \infty}x_n=b$ and $x_0=a$, this should be extended to $$|a-b|\leq \sum_{i=1}^\infty|x_{i-1}-x_{i}|$$
Is it right? Is not a surprising idea, but I think I could not find this usually on textbooks.
Your result is correct. You can formalize the proof as follows: For every $n \in \Bbb N$ is $$ |a-b| = |(a-x_1)+ (x_1-x_2) + \cdots + (x_{n-1}-x_n) + (x_n - b)| \\ \le \left(\sum_{i=1}^n |x_{i-1}- x_i|\right) + |x_n - b| \, , $$ using the triangle inequality for finitely many terms.
If the series $\sum_{i=1}^\infty |x_{i-1}- x_i|$ converges then we can take the limit $n \to \infty$ and conclude that $$ |a-b| \le \sum_{i=1}^\infty |x_{i-1}- x_i| $$ since $\lim_{n\to \infty}x_n=b$.
Otherwise $\sum_{i=1}^\infty |x_{i-1}- x_i|$ diverges to $+\infty$ and the same estimate holds (trivially) in the set of extended real numbers: $|a-b| \le +\infty$.
