If $(x_n)$ is an unbounded increasing sequence then $\sum (1-x_n/x_{n+1})$ diverges Let {$x_n$} be monotone increasing sequence of positive real numbers. Show that if {$x_n$} is unbounded, then $\sum_{n=1}^{\infty}(1-\frac{x_n}{x_{n+1}})$ diverges. 
 A: Using the well-known inequality $\ln a\leq a-1$ with $a=x_{k+1}/x_k$ we obtain
$$
\ln x_{k+1}-\ln x_k\leq \frac{x_{k+1}-x_k}{x_k}
$$
Adding these inequalities for $k\in\{1,2,\dots,n-1\}$, we obtain
$$
\ln x_{n}-\ln x_1\leq\sum_{k=1}^{n-1} \frac{x_{k+1}-x_k}{x_k}
$$
but the sequence $\{x_n\}$ in increasing and unbounded, hence
$$
\sum_{k=1}^{\infty} \frac{x_{k+1}-x_k}{x_k}=+\infty.\tag{1}
$$
Now  suppose, for contradiction, that the  considered series does converge, $i.e.$
$$\sum\limits_{k=1}^{\infty} \dfrac{x_{k+1}-x_k}{x_{k+1}}<+\infty\tag{2}$$
 then its general term must tend to zero,
that is $\lim\limits_{k\to\infty}\dfrac{x_k}{x_{n+1}}=1$, which implies that
$$
\dfrac{x_{k+1}-x_k}{x_{k+1}}\sim_{\infty}\dfrac{x_{k+1}-x_k}{x_{k}}
$$
and contradicts (1). This contradiction proves that (2) is absurd and the series
$\sum\limits_{k=1}^{\infty} \dfrac{x_{k+1}-x_k}{x_{k+1}}$  must be divergent.
A: If $\frac{x_n}{x_{n+1}}$ does not tend to $1$ then the sequence diverges because the $n$th term does not tend to zero.  If $\frac{x_n}{x_{n+1}}\to 1$ then your sequence is equivalent to the sequence $\log(\frac{x_n}{x_{n+1}})$ which diverges by a telescoping series argument.  The reason the sequence is equivalent to $\log(\frac{x_n}{x_{n+1}})$ is because as $x\to 0$ one has $\log (1+x)\sim x$.
