Suppose $f(x)>0$ and $f$ is continuous on $[0,\infty)$ and $$\lim_{x\rightarrow\infty}\frac{f(x+1)}{f(x)}<1$$ How to see that $\int^\infty_0f(x)dx$ converges?
I think I should use definition. $\forall \epsilon>0$, there exists $M>0$ s.t. $|\frac{f(x+1)}{f(x)}-L|<\epsilon$ for all $x>M$ where $L<1$. But how to proceed from here to show that $\lim_{R\rightarrow\infty}\int^R_0f(x)dx$ exists? It should be easy, but haven't got the key idea.
Let the limit be $L \in [0,1)$. Let $\epsilon>0$ be small enough so that again $a:= L + \epsilon \in [0,1)$. Then by the given convergence, $\exists C>0$ such that $\forall x \geq C$
$$ f(x+1) < (L+ \epsilon)f(x) = a f(x).$$
Let $n_0$ denote the first integer after $C$. Then by induction we have that $\forall k \in \mathbb{N}$
$$f(n_0 +k) \leq a^kf(n_0).$$
Hence, by a change of variables, we have $\forall k \in \mathbb{N}$ that
$$\int_{n_0+k}^{n_0+(k+1)} f(x) dx = \int_{n}^{n+1} f(x+k) dx \leq a^k \int_{n}^{n+1} f(x) dx.$$
Therefore
$$\int_{n_0}^{\infty} f(x) dx \leq \int_{n_0}^{n_0+1} f(x) dx \sum_{k=0}^{\infty} a^k < \infty,$$
where the first inequality is justified by the Monotone Convergence Theorem.
Since $f$ is continuous on the compact set $[0,n_0]$, it is also integrable there.