The test says:

Given two functions $f$ and $g$ defined positive and integrable on $\left [ a,+\infty \right [$. Suppose that $\lim_{x\to \infty}\frac{f(x)}{g(x)}=L$ exists.

If $0< L < \infty$: $\int_{a}^{\infty }g(x)dx$ converges $\Leftrightarrow \int_{a}^{\infty }f(x)dx$ converges

If $L=0$: $\int_{a}^{\infty }g(x)dx$ converges $\Rightarrow \int_{a}^{\infty }f(x)dx$ converges

The proof goes by using the definition of the limit at infinity by setting an $\epsilon $ and stating the existence of a certain $M$ above which $\frac{f(x)}{g(x)}$ is within $\epsilon $ of $L$ and then using the comparison theorem to deduce the convergence of the integral of a function by its upper boundedness by a function whose integral converges. Here is a link if I explained badly (from the end of page 1 to the top of page 2): http://www.math.toronto.edu/~alfonso/137/137_1516_LCT.pdf


I fail to see why this proof fails in the left implication direction when $L=0$? We would have $-\epsilon g(x)\leq f(x)\leq \epsilon g(x)$ so why would convergence of $\int_{a}^{\infty }f(x)dx$ not imply convergence of $\int_{a}^{\infty }g(x)dx$?

Thanks very much in advance.

  • 1
    $\begingroup$ Notice that the Limit comparison test does not assume that $0\le f(x)\le g(x)$, and that the first statement should say if $0<L<\infty$, then $\int_a^{\infty}f(x)dx$ converges $\iff$ $\int_a^{\infty}g(x)dx$ converges. $\endgroup$ – user84413 Aug 28 '16 at 22:14
  • $\begingroup$ @user84413 Thank you! corrected it $\endgroup$ – John11 Aug 28 '16 at 22:19

A basic counterexample would be $f(x)=\frac 1{x^2}$ and $g(x)=\frac 1x$. $\lim_{x\to\inf}\frac{f(x)}{g(x)}=0$, and $\int_{a>0}^\infty f(x)dx$ converges, but $\int_{a>0}^\infty g(x)dx$ does not. The reason the proof breaks down can be understood by looking at the inequality you gave.

$$x>M\implies-\epsilon g(x)\le f(x)\le\epsilon g(x)$$

The right inequality is a perfect fine for comparison test, since $0\le f(x)\le \epsilon g(x)$, but since the $f$, $g$, and $\epsilon$ are nonegative, the left side is trivially true, and is useless for a comparison. More specifically, the inequality $-\epsilon g(x)\le 0\le f(x) $ does not squeeze $g(x)$ between two functions with the same limit. That is precisely why the proof uses $\epsilon=\frac L2$, assuring that the left inequality gives the useful comparison $\frac L2g(x)\le f(x)\le 0$. When $L=0$, the argument no longer works. In fact, no argument would work, as this would contradict the counterexamples.

  • $\begingroup$ +1 Perfect explanation! Minor note, I think you may have swapped $f(x)$ and g(x)$ while writing the explanation so I corrected it. $\endgroup$ – John11 Aug 28 '16 at 22:47
  • $\begingroup$ Thanks for pointing that out. was thinking $\ge$ but writing $\le$. $\endgroup$ – Kajelad Aug 28 '16 at 23:02

Let $f(x)=\frac{1}{x^2}$ and $g(x)=\frac{1}{x}$.

Then $\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{1/x^2}{1/x}=\lim_{x\to\infty}\frac{1}{x}=0$ and $\int_1^{\infty}f(x)dx$ converges, but $\int_1^{\infty}g(x)dx$ diverges.


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