Let $f:[0,\infty)\to \mathbb R$ be Continuous and strictly decreasing function such that : $\lim_{x\to \infty}f(x)=0$

Question

Let $f:[0,\infty)\to \mathbb R$ be Continuous and strictly decreasing function such that : $\lim_{x\to \infty}f(x)=0$ prove $$\int_0^\infty \frac{f(x)-f(x+1)}{f(x)}$$ Diverges.

I can not prove. please help .

Answer 1

For any $n\in\mathbb{N}^+$, let $$E_n = \left\{x\in\mathbb{R}^+ : \frac{1}{n+1}\,f(x) \leq f(x+1) \leq \frac{1}{n}\, f(x)\right\}. $$ Since $f$ is decreasing and positive, $\bigcup_{n\geq 1}E_n = \mathbb{R}^+$. We have: $$ \left(1-\frac{1}{n+1}\right)\mu(E_n)\geq\int_{E_n}\frac{f(x)-f(x+1)}{f(x)}\,dx \geq \left(1-\frac{1}{n}\right) \mu(E_n)$$ hence in order to ensure the convergence of the given integral both the following constraints have to hold: $\sum_{n\geq 1}\mu(E_n)=+\infty$ and $\sum_{n\geq 1}\left(1-\frac{1}{n}\right)\mu(E_n)<+\infty$. By the asymptotic comparison criterion for non-negative series, these constraints are not compatible, hence the given integral is divergent.