Q:

Given $f$ differentiable & nonzero on $[1, \infty)$, $\lim_{x\to\infty} \frac{x f'(x)}{f(x)} < -1$ and non-infinite. Show $\int_1^\infty f$ converges

We know that $1 + \frac{x f'}{f} < 0$.
I'm unsure how to proceed but guessing the proof involves knowing that:
If $\lim_{x \to b} \frac{f(x)}{g(x)}$ exists and is greater than 0, then $\int_a^b g(x)$ converges if and only if $\int_a^b f(x)$ converges.
If $\lim_{x \to b^-} (f(x))^\frac{1}{x}$ exists in the extended reals, then $\int_a^b f(x)$ converges if the limit is less than 1 and diverges otherwise.

A:

Since $f$ is continuous and non zero, then $f$ is positive or negative.
Assume that $f$ is positive.
By assumption, there exist numbers $\alpha < -1$ and $K \geq 1$ such that
$$
\frac{x f'(x)}{f(x)} \leq \alpha \qquad \forall x\geq K,
$$ 
i.e.
$$
\frac{d}{dx} \log f(x) = \frac{f'(x)}{f(x)} \leq \frac{\alpha}{x}
\qquad \forall x\geq K.
$$
Hence
$$
\log \frac{f(x)}{f(K)} \leq \alpha \log \frac{x}{K},
\qquad\forall x\geq K,
$$
i.e. $f(x) \leq C x^{\alpha}$ for every $x\geq K$, for some constant $C>0$.
Since $f\in C([1,\infty))$ and $\alpha < -1$, this implies that $\int_1^\infty f < \infty$.
A similar argument works also in the case $f<0$.

