Prove that $\sum_{n=1}^{\infty}\left (1-\frac{a_{n}}{a_{n+1}}\right)$ converges Let $a_n >0$ for all $n\in N$. Suppose that $(a_n)$ is increasing and bounded. Prove that $$\sum\limits_{n=1}^{\infty} \left(1-\frac{a_{n}}{a_{n+1}}\right)$$ converges.
I've been trying the Abel's Test but I can't get the conclusion. Anyone help me please?
 A: Let $M$ be a lower bound of $(a_n)$ and $M'$ be an upper bound. Partial sum:
$$S_n = \sum_{r=1}^n \frac{a_{r+1} - a_r}{a_{r+1}} \le \frac1M \sum_{r=1}^n a_{r+1} - a_r = \frac1M (a_{n+1} - a_1) \le \frac1M (M' - a_1)$$
So $(S_n)$ is bounded. But $(S_n)$ is increasing (since its general term is positive). Therefore $(S_n)$ is convergent.
A: Two useful generalities :
$$\text {(1). If } b_n\geq 0 \text { then } \sum_nb_n<\infty \iff \prod_n(1+b_n)<\infty.$$
$$\text {(2). If  } 0<b_n<1 \text{ then } \sum_nb_n<\infty \iff \prod_n(1-b_n)>0.$$
Neither of these is difficult.  We  will use (2).
For $0<a_n\leq a_{n+1}<M:$ 
(I). If $\max_n a_n$ exists then $1-a_n/a_{n+1}=0$ for all but finitely many $n,$ and we are done.
(II). If $\max_n a_n$ does not exist, then there is a strictly increasing $f:\mathbb N\to \mathbb N$ such that $$f(1)=1$$  $$\text {and } \quad a_{f(n)}<a_{f(n+1)}$$  $$ \text { and } \quad f(n)\leq j<f(n+1)\implies a_j=a_{j+1}.$$
We have $\sum_j1-\frac   {a_j}  {a_{j+1}}  =\sum_n1-\frac {a_{f(n)}}{a_{f(n+1)}}.$
Because for each $j$ there is (unique) $n_j$ such that $f(n_j)\leq j<f(n_j+1)$. Therefore  
(i). If $j+1<f(n_j+1)$ then  $a_j=a_{j+1}.$
(ii). If $j+1=f(n_j+1)$ then, since $f(n_j)\leq j<f(n_j+1)$ , we have $a_j=a_{f(n_j)}.$
Now let $ b_n=1-a_{f(n)}/a_{f(n+1)}.$ We have $0<b_n<1.$ We have $$\prod_{j=1}^n(1-b_j)=   \prod_{j=1}^n \frac {a_{f(j)}}{a_{f(j+1)}}=\frac {a_{f(1)}}{a_{f(n+1)}}=\frac {a_1}{a_{f(n+1)}}>a_1/M>0.$$ By (2), we are done.
