Let $f:[0,\infty)\rightarrow \mathbb{R}$ be a continuous and strictly decreasing function. If $\displaystyle\lim_{x\rightarrow\infty}f(x)=0$, prove that the following integral is divergent : $$\int_0^\infty \frac{f(x)-f(x+1)}{f(x)}dx$$
First, it's not hard to see that $f(0)>0$ and also $f(x)>0$ for each $x$.
I think we have to show that behavior of $\frac{f(x)-f(x+1)}{f(x)}$ is "similar" to $\frac1{x+1}$ or $\frac{f(x+1)}{f(x)}$ is "similar" to $\frac{x}{x+1}$ for big values of $x$.
But meaning of "similar" is not clear yet !
The most naive and rough approach is to show $\frac{f(x+1)}{f(x)}<1-k$ for some $0<k<1$ and therefore $f(x)-f(x+1)>kf(x)$.
One step forward, it's good to compare $\frac{f(x+1)}{f(x)}$ and a multiple of $\frac{x}{x+1}$ or equivalently to compare $C (x+1)f(x+1)$ and $xf(x)$. But still I can't extend this idea .
I tried more ways but I couldn't find a proper way to solve the problem.
Any hint is appreciated !
Let $n\in \mathbb{N}$. For every $m\in \mathbb{N}$, we have
\begin{align} \int_n^{n+m} \frac{f(x) - f(x+1)}{f(x)}\,dx &= \sum_{k=0}^{m-1} \int_{n+k}^{n+k+1} \frac{f(x) - f(x+1)}{f(x)}\,dx\\ &= \sum_{k=0}^{m-1} \int_0^1 \frac{f(n+k+t) - f(n+k+1+t)}{f(n+k+t)}\,dt\\ &= \int_0^1 \sum_{k=0}^{m-1} \frac{f(n+k+t)-f(n+k+1+t)}{f(n+k+t)}\,dt. \end{align}
Since $f$ is positive and decreasing, we have
$$\frac{f(n+k+t) - f(n+k+1+t)}{f(n+k+t)} \geqslant \frac{f(n+k+t)-f(n+k+1+t)}{f(n+t)}$$
for all $k\in\mathbb{N}$ and all $t\in [0,1]$, hence
\begin{align} \sum_{k=0}^{m-1} \frac{f(n+k+t)-f(n+k+1+t)}{f(n+k+t)} &\geqslant \sum_{k=0}^{m-1} \frac{f(n+k+t) - f(n+k+1+t)}{f(n+t)}\\ &= \frac{f(n+t) - f(n+m+t)}{f(n+t)}. \end{align}
Now choose $m$ so large that
$$f(n+m) \leqslant \frac{1}{2}\cdot f(n+1),$$
then we have
$$\int_n^{n+m} \frac{f(x)-f(x+1)}{f(x)}\,dx \geqslant \int_0^1 \frac{f(n+t) - f(n+m+t)}{f(n+t)}\,dt \geqslant \int_0^1 \frac{1}{2}\,dt = \frac{1}{2}.$$
But if the integral were convergent, then for every $\varepsilon > 0$ there would exist an $n\in\mathbb{N}$ such that
$$\int_n^\infty \frac{f(x)-f(x+1)}{f(x)}\,dx < \varepsilon.$$
By the above, such an $n$ does not exist for $\varepsilon \leqslant \frac{1}{2}$, hence the integral is divergent.