Prove that $\int_0^\infty \frac{f(x) - f(x + 1)}{f(x)}\, dx$ diverges This is one of the problems I got in a test today, and one that i got no idea how to solve.

Let $f:[0,\infty) \to \mathbb{R}$ be strictly decreasing continuous function with $\displaystyle\lim_{x\to \infty} f(x) = 0$. Prove that $$ \int_0^\infty \frac{f(x) - f(x + 1)}{f(x)}\, dx$$ diverges.

I tries using the fact that $f$ has maximum and minimum on every closed intervals $[0,b]$  then limitting $b\to \infty$, $$\int_0^b \left( 1 - \frac{f(x + 1)}{f(x)}\right) dx  = b - \int_0^b \frac{f(x + 1)}{f(x)} dx \geq b - \int_0^b  f(0)/f_{\min}$$

But the integral on the right doesn't seem to converge as $b\to\infty$.

Any hint?
(done on my phone)

If $f'$ exists then
$f(x+1)-f(x) \approx f'(x)$
so the integrand is about
$-f'(x)/f(x) = (-\ln(f(x)))'$
so the integral out to $T$
is about $-\ln(f(T))
\to \infty$.

Not sure how to modify this
for general $f$.
(done on my phone)

If $f'$ exists then
$f(x+1)-f(x) \approx f'(x)$
so the integrand is about
$-f'(x)/f(x) = (-\ln(f(x)))'$
so the integral out to $T$
is about $-\ln(f(T))
\to \infty$.

Not sure how to modify this
for general $f$.