For $(x_n)$ increasing, $\sum_{n=1}^{\infty}\left(1-\frac{x_n}{x_{n+1}}\right)$ if $(x_n)$ is bounded and diverges if it is unbounded 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.

I try to rewrite the proof cleaning some details.
Call $r_n=\frac{x_n}{x_{n+1}}\in(0,1]$. If there's an infinite number of terms such that $1-r_n=1-\frac{x_n}{x_{n+1}}\geq \frac12$ then the series clearly diverges, while by rewriting the condition as $x_{n+1}\geq 2 x_n$ also the sequence diverges.
If instead eventually $\frac12\leq r_n\leq 1$, then we can use that 
$$\log r\leq r-1\leq C\log r\qquad\text{for}\quad \frac12 \leq r\leq 1$$
for some positive $C$ to obtain
$$\sum_{n=1}^N\left(\frac{x_n}{x_{n+1}}-1\right)\leq C \sum_{n=1}^N \log \frac{x_n}{x_{n+1}}=C\log\frac{x_1}{x_N}$$
and
$$\sum_{n=1}^N\left(\frac{x_n}{x_{n+1}}-1\right)\geq \sum_{n=1}^N \log \frac{x_n}{x_{n+1}}=\log\frac{x_1}{x_N}.$$
Therefore the series has the same behaviour of $\lim_{N\to\infty}\log\frac{x_1}{x_{N}}$, which diverges iff $x_N\to\infty$.
