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Let $a_n >0$ for all $n\in N$. Suppose that $(a_n)$ is increasing and bounded. Prove that $\displaystyle\sum\limits_{n=1}^{\infty} \left(1-\dfrac{a_{n}}{a_{n+1}}\right)$ converges.

I've been trying the Abel's Test but I can't get the conclusion. Anyone help me please?

  • 0
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2 Answers 2

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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.

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Two useful generalities :

$$\text {(1). If } b_n\geq 0 \text { then } \sum_nb_n<\infty \iff \prod_n(1+b_n)<\infty.$$

$$\text {(2). If } 00.$$

Neither of these is difficult. We will use (2).

For $0

(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)}

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

(i). If $j+1

(ii). If $j+1=f(n_j+1)$ then, since $f(n_j)\leq j

Now let $ b_n=1-a_{f(n)}/a_{f(n+1)}.$ We have $0a_1/M>0.$$ By (2), we are done.