# Convergent series $\sum_n\frac{a_{n+1}-a_n}{a_n}$ implies that $(a_n)$ is bounded [duplicate]

Suppose we have a sequence $$(a_n)_{n\in\mathbb N}$$ with $$a_n>0$$ and $$a_{n+1}\geq a_n$$ for all $$n\in\mathbb N$$.

Then, I want to prove that $$\sum_{n=1}^\infty\left(\frac{a_{n+1}-a_n}{a_n}\right)\text{ converges}\quad\implies\quad (a_n)_{n\in\mathbb N}\text{ is bounded}.$$

I tried to prove this by contradiction and using inequalities, but I never arrived at a divergent series.

We use the easy result that if $$c_n \ge 0, \Sigma{c_n} < \infty$$ if and only if $$\Sigma{\log (1+c_n)} < \infty$$ with $$c_n=(\frac{a_{n+1}}{a_n}-1)$$ which imediately implies $$\log{a_n}$$, hence $$a_n$$ bounded
($$\frac{x}{2} \le \log(1+x) \le x, 0 \le x \le 1$$ and each convergence in the iff above, implies $$c_n \le 1$$ eventually)
By the AM-GM inequality, for $$m>n$$ $$\frac{a_m}{a_{m-1}}+\frac{a_{m-1}}{a_{m-2}}+\ldots +\frac{a_{n+1}}{a_{n}}\ge (m-n)\cdot \sqrt[m-n]{\frac{a_m}{a_{m-1}}\frac{a_{m-1}}{a_{m-2}}\cdots\frac{a_{n+2}}{a_{n+1}}} =(m-n)\cdot\sqrt[m-n]{\frac{a_m}{a_{n+1}}}$$ and so with $$S_n:=\sum_{k=1}^{n-1}\frac{a_{k+1}-a_k}{a_k}$$, $$\tag1S_m-S_n\ge (m-n)\left(\sqrt[m-n]{\frac{a_m}{a_{n+1}}}-1\right)$$ As the sequence of the $$S_n$$ is Cauchy (and non-decreasing), there exists $$n$$ such that $$0\le S_m-S_n<1$$ for all $$m>n$$. Using this and solving $$(1)$$ for $$a_m$$, $$\left(1+\frac1{m-n}\right)^{m-n}a_{n+1}\ge a_m.$$ We know that $$\lim_{k\to\infty}\left(1+\frac1{k}\right)^{k}= e$$, hence $$s:=\sup\left\{\,\left(1+\frac1{k}\right)^{k}\Bigm|k\in\Bbb N\right\}$$ is finite and $$a_k\le \max\{a_1,a_2,\ldots,a_n,sa_{n+1}\}$$ for all $$k\in\Bbb N$$.