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

Question

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.

Answer 1

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)

Answer 2

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 $k\in\Bbb N$.