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
MY QUESTION:
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.
Best Answer
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.