Ambiguity in the limit comparison test for integrability Check out theorem 17.3 in the following link:
http://home.iitk.ac.in/~psraj/mth101/lecture_notes/lecture18.pdf
I have issues understanding its last sentence: in case c=0 and ∫g(x) convergent, what's the difference with the situation c≠0, c≠∞ and ∫g(x) convergent? I don't get what this "convergence of ∫g(x) implies convergence of ∫f(x)" means, seems to me like those two situations are the same and the case c=0 is irrelevant...though I'm sure it's wrong and there's something there I don't get!
Thanks a lot in advance for your answers, I appreciate it.
Julien.
 A: The statement $\int g$ converges implies $\int f$ converges is different from the statement $\int g$ converges if and only if $\int f$ converges. In the former case you can have $\int f$ converge and not $\int g$, just not the other way around. In the latter statement the implication goes both ways the first it only goes one way.  That's what they mean by "both" converge or "both" diverge together.
A: Let $c = \lim_{t \to \infty} \frac{f(t)}{g(t)}.$ Since $f$ and $g$ are nonnegative we have $c \geqslant 0$. 
If $0 < c < \infty$ then for $t$ sufficiently large and $\epsilon < c$ we have 
$$0 < c - \epsilon < \frac{f(t)}{g(t)} < c + \epsilon \\ \implies(c- \epsilon)g(t) < f(t) < (c+\epsilon)g(t).$$
The left inequality can be used to show that divergence of $\int g$ implies divergence of $\int f.$ The right inequality can be used to show that convergence of $\int g$ implies convergence of $\int f.$  Hence we say the integrals converge and diverge together, or one converges if and only if the other converges.
If $c = 0$ then we have for sufficiently large $t$ and some $\epsilon > 0$
$$- \epsilon < \frac{f(t)}{g(t)} < \epsilon \\ \implies -\epsilon g(t) < f(t) < \epsilon g(t).$$
The right inequality can be used to show that convergence of $\int g$ implies convergence of $\int f.$  The left inequality is useless because $- \epsilon < 0$. Here we could have $\int g$ diverge and $-\epsilon \int g \to -\infty,$ but the integral $\int f$ could still be convergent.  Hence, we have only a one-directional implication in this case. 
