How to a function converges or diverges by comparison test?

Question

Question: $\int_{4}^{\infty}f(x)dx$ is an improper integral, where $$f(x)=\frac{3x+8}{x^4+7}$$

For the comparison test we have to find a comparison function, say $g(x)$ and

• for any $f(x)>g(x)$, if $f(x)$ converges, then $g(x)$ also converges;
• for any $f(x)<g(x)$, if $f(x)$ converges, we can't guarantee whether $g(x)$ converges or not.

For this question I have got $g(x)=\frac{1}{x^3}$, which converges but is smaller than the given $f(x)$.

Now, how can I determine its [f(x)] convergence from comparison test?

I read somewhere that in these cases I have to find a new value of $g(x)$ which converges and greater than $f(x)$ (which I am not sure how to find). Also, does that mean that the convergence or divergence will depend on the chosen value of $g(x)$?

Answer 1

We can refer indeed to limit comparison test for integrals, since eventually for some $0<\varepsilon<3$

$$\lim_{x\to \infty} \frac{f(x)}{g(x)}=\lim_{x\to \infty} \frac{3x^4+8x^3}{x^4+7} = 3 \iff (3-\varepsilon)g(x)\le f(x)\le (3+\varepsilon)g(x)$$

this allow to refer to direct comparison test with $(3+\varepsilon)g(x)$.

More in general, for improper integrals at infinity, with $f(x)$ and $g(x)$ (at least eventually) positive, the condition

$$\lim_{x\to \infty} \frac{f(x)}{g(x)}=L\in\mathbb R$$

suffices to claim that both related integrals diverges or converges and the argument also works for the extreme cases when $L=0$ with $g(x)$ convergent and when $L=\infty$ with $g(x)$ dinvergent.

Answer 2

An alternative solution:

$$f(x)=\frac{3x+8}{x^4+7}\le \frac{3x+2x}{x^4+7}\le \frac{5x}{x^4}$$ The function $g(x)=\frac{5}{x^3}$ meets our requirements.

Answer 3

Since the denominator is never equal to $0$, you only have to study the behaviour of the function as $x\to +\infty$: $$\int_4^{+\infty}\frac{3x+8}{x^4+7}dx \sim \int_4^{+\infty}\frac{3x}{x^4}dx \sim \int_4^{+\infty}\frac{x}{x^4}dx=\int_4^{+\infty}\frac{dx}{x^3}$$ Since $3>1$, the integral converges.

Answer 4

In order to avoid smart estimates we can use $${3x+8\over x^4+7}={3x\over x^4+7}+{8\over x^4+7}\\ \le {3x\over x^4}+{8\over x^4}={3\over x^3}+{8\over x^4}$$

Answer 5

The comparison test is really the squeeze theorem with $f(x) \equiv 0$ as one of the function. So don't forget that you need to establish that both the functions are non-negative.

Generally, once you have a function in one direction, it tends to not be difficult to find one in the other direction. One tactic that often works is to simply add an order of magnitude (it's generally overkill, but that's okay). In this case, that gives you $\frac{10}{x^3}$. Multiply top and bottom by $x$ to get $\frac{10x}{x^4}$. Now, if $x > 4$, then $3x+8<5x<10x$, so $f(x)<\frac{10x}{x^4+7}<\frac{10x}{x^4}$.