According to the Limit Comparison Test, if the improper integral of the quotient of two functions converges, the improper integrals of these two functions should either both converge or both diverge.
Suppose I want to calculate $\int_1^\infty \frac{\sqrt{x+1}}{x^2}dx$. If I were to use the Limit Comparison Test, I would treat $\frac{\sqrt{x+1}}{x^2}$ as the numerator, or $f(x)$, and I would treat $\sqrt{x+1}$ as the denominator, or $g(x)$. Dividing the two results in $\frac{1}{x^2}$, and the integral of that from $1$ to $\infty$ would converge, by the $p$-test. But the integral of $g(x)$ diverges on that range.
By that logic, we can expect the integral of $f(x)$ to diverge as well. However, $\int_1^\infty \frac{\sqrt{x+1}}{x^2}$ converges, according to the online integral calculator. So what went wrong here?
Best Answer
This is not true. You don't take the improper integral of a quotient of two functions. That is not what the Limit Comparison Test states. It states
"If $f(x),g(x) > 0$ and are continuous on $[a,\infty)$ and $0 < \displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)} < \infty$, then $\displaystyle\int_{a}^{\infty}f(x)dx$ and $\displaystyle\int_{a}^{\infty}g(x)dx$ both converge or both diverge."
This choice of $g(x)$ does not work for LCT. Yes, you would get $\dfrac{1}{x^2}$, but $\displaystyle\lim_{x\to\infty}\frac{1}{x^2} = 0$, rendering LCT useless.
(Answer) Assuming we don't know if the given integral in question converges or diverges, we know $\dfrac{\sqrt{x+1}}{x^2} > 0$ and is continuous on $[1,\infty)$. Similarly, $\dfrac{1}{x^{3/2}} > 0$ and is continuous on $[1,\infty)$. Then $\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{x+1}/x^2}{x^{-3/2}} = 1$, and $0<1<\infty$. We see that $\displaystyle\int_{1}^{\infty}x^{-3/2}dx$ converges by the p-test. Therefore by LCT, we conclude $\displaystyle\int_{1}^{\infty}\frac{\sqrt{x+1}}{x^{2}}dx$ converges as well.
Please let me know if there are any confusions.