Let $\{x\_n\}$ be monotone increasing sequence of positive real numbers. Show that if $\{x\_n\}$ is bounded, then $\sum\_{n=1}^{\infty}\left(1-\frac{x\_n}{x\_{n+1}}\right)$ converges. On the other hand, if the sequence is unbounded, the series is divergent.

The following is the proof given by my lecturer, but I do not understand the proof:

Let $u\_m = x\_{m+1} -x\_m$ and $d\_m = \sum\_{k=1}^m u\_k = x\_{m+1} - x\_1$ and hence $x\_{m+1} = d\_m + x\_1$

$$\sum\_{n=1}^{\infty}\left(1-\frac{x\_n}{x\_{n+1}}\right) = \sum\_{n=1}^{\infty}\frac{u\_m}{d\_m +x\_1} $$

Then he says that if $d\_m$ diverges, then the sum diverges. If $d\_m$ converges, then then sum converges.

He says that it has something to do with Gauss Test.

Can someone explain the proof to me. Or does anyone has a better proof.