Need a proof for a generalized sequence test. Suppose $f(x)>0 $, $f$ is decreasing on $[1,\infty)$.  $g(x),h(x)$ are both increasing on $[1,\infty)$.And:
$$g(x)>h(x);\quad h(x)\xrightarrow[]{x\to+\infty}+\infty$$
We denote that $$r=\lim_{x\to \infty}\dfrac{g'(x)f(g(x))}{h'(x)f(h(x))}$$
(Here, we just consider the case that the limit exists or $r=\infty$.)$$$$
The question is :$\boxed{\text{Discuss whether $\int_1^\infty f(x)dx$ is convergent or not ,when $r\neq1$.}}$
$$$$Personally, I think the conclusion is of great significance for the test of a sequence. For example, we assume$$g(x)=x+1, h(x)=x,a_n=f(n)$$
Then if $$r=\frac{a_{n+1}}{a_n}<1$$$$\sum_{n=1}^\infty a_n\text{ is convergent.}$$
Obviously the sum is divergent, when $r>1$.$$$$
We can also assume $g(x)=2x \text{ , } x^2 \text{ or } x^3 \text{ ,etc. while } h(x)=x $
$$$$If we assume that $g(x)=e^x\text{ ,while }h(x)=x$, is it called "Markov test"?
 A: Indeed, $r < 1$ implies convergent and $r > 1$ implies divergent. We give the proof that $r < 1$ implies convergent; by barely altering it (change "$\le$" to "$\ge$"), you get $r > 1$ implies divergent.
Fix $\alpha \in (r,1)$. Take $y_0 > 1$ such that $f(g(y))g'(y) \le \alpha f(h(y))h'(y)$ for all $y \ge y_0$. Let $y_1 > y_0$ be such that $h(y_1) = g(y_0)$ (existence due to $g > h$ and $h \to \infty$; uniqueness due to $h$ increasing). Then let $y_2$ be such that $h(y_2) = g(y_1)$. Let $y_3,y_4,\dots$ be defined similarly. Note that $y_n \to +\infty$; indeed, otherwise the $y_n$'s have a limit, say $y$, but then we'd have $g(y) = h(y)$, contradicting $g > h$. Therefore, $$\int_{y_0}^\infty f(g(y))g'(y)dy = \sum_{n=0}^\infty \int_{y_n}^{y_{n+1}} f(g(y))g'(y)dy.$$ Note that, for $n \ge 1$, a simple change of variables gives $$\int_{y_n}^{y_{n+1}} f(h(y))h'(y)dy = \int_{y_{n-1}}^{y_n} f(g(y))g'(y)dy,$$ so induction gives $$\int_{y_n}^{y_{n+1}} f(g(y))g'(y)dy \le \alpha^n \int_{y_0}^{y_1} f(g(y))g'(y)dy$$ for each $n \ge 0$. Since $\alpha < 1$, we see that $$\int_{g(y_0)}^\infty f(x)dx = \int_{y_0}^\infty f(g(y))g'(y)dy < \infty.$$ This shows $\int_1^\infty f(x)dx$ is convergent. 
