# Prove limit comparison test for Integrals

• CGandC
In summary: L2 |g(x)| in (R, \infty) and it follows that\int_R^\infty |f(x)|\,dx converges if and only if \int_R^\infty |g(x)|\,dx converges.
CGandC
Homework Statement
Theorem: Given the continuous functions ## f,g : [ 1, +\infty ) \to \mathbb{R} ## that satisfy ##\lim _{x \rightarrow+\infty} \frac{f(x)}{g(x)}=L>0##. Prove the improper integrals ## \int_1^{+\infty} f(x)dx , \int_1^{+\infty} g(x)dx ## converge or diverge together.
Relevant Equations
* ( Integral comparison ) Let ## f,g : [ a,\omega) \to \mathbb{R} ## ( where ## \omega \in \mathbb{R} ## or ## \omega = \infty ## ). Suppose for all ## b < \omega ## it occurs that ## f,g \in \mathbb{R([a,b])} ##. Suppose ## 0 \leq f \leq g ##. If ## \int_a^\omega g < \infty ## then also ## \int_a^\omega f < \infty ##, and if ## \int_a^\omega f = \infty ## then ## \int_a^\omega g = \infty ## .

* ## f \in R([ a,b]) ## means ## f ## is Riemann-Integrable on ## [ a,b] ##
Attempt:
Note we must have that
## f>0 ## and ## g>0 ## from some place
or
## f<0 ## and ## g<0 ## from some place
or
## g ,f ## have the same sign in ## [ 1, +\infty) ##.

Otherwise, we'd have that there are infinitely many ##x's ## where ##g,f ## differ and sign so we can chose a sequence ## 1<x_n \to \infty ## s.t. ## \lim _{n \rightarrow+\infty} \frac{f(x_n)} {g(x_n)}<0## , a contradiction to the given ##\lim _{x \rightarrow+\infty} \frac{f(x)}{g(x)}=L>0 ##.

Suppose ## f>0 ## and ## g>0 ## from some place ( proof for ## f<0 ## and ## g < 0 ## from someplace or proof for the case where ## g,f ## have same sign in ## [1, +\infty ) ## are very similar
).
From the given ## \lim _{x \rightarrow+\infty} \frac{f(x)}{g(x)}=L>0 ## we can say that ## \forall \epsilon>0 . \exists R>0 . \forall x>R . | \frac{f(x)}{g(x)} - L | < \epsilon ##.
Note that ## -\epsilon < \frac{f(x)}{g(x)} - L < \epsilon \longrightarrow
L - \epsilon < \frac{f(x)}{g(x)} < L + \epsilon \longrightarrow (L-\epsilon)g(x) < f(x) ##.

Hence, for ## \epsilon = \frac{L}{2} ## there exists ## R>0 ## s.t. for all ## x>R ## we have that ## (L-\epsilon)g(x) = \frac{L}{2} g(x) < f(x) ##.
Suppose without loss of generality that ## R>1 ##.
Suppose that ## \int_1^{+\infty} f(x)dx ## converges and we'll show that ## \int_1^{+\infty} g(x)dx ## converges ( proving that if ##\int_1^{+\infty} g(x)dx## converges then ## \int_1^{+\infty} f(x)dx ## converges is similar if we look at ## f(x) < ( L + \epsilon) g(x) ## ) .
Under the new assumption, note that ## \int_R^{+\infty} f(x)dx > \frac{L}{2} \int_R^{+\infty} g(x) dx ##, hence from comparison test for integrals ## \int_R^{+\infty} g(x) dx ## exists.
Now, note that ## \int_1^{+\infty} g(x)dx = \int_1^{R} g(x)dx + \int_R^{+\infty} g(x)dx ##, Hence ## \int_R^{+\infty} g(x)dx = \int_1^{+\infty} g(x)dx - \int_1^{R} g(x)dx ## and since ## \int_R^{+\infty} g(x) dx ## exists then these two integrals also exist, hence ## \int_1^{+\infty} g(x)dx ## exists.

( After proving that if ##\int_1^{+\infty} g(x)dx## converges then ## \int_1^{+\infty} f(x)dx ## converges, we proved that ##\int_1^{+\infty} g(x)dx ~ \text{ converges} \iff \int_1^{+\infty} f(x)dx ~ \text{converges} ##. Hence we also proved ##\int_1^{+\infty} g(x)dx ~\text{ diverges} \iff \int_1^{+\infty} f(x)dx ~ \text{diverges} ## )
## \square ##There are couple of things that I have questions about:
1. Where have I used the fact that ##f,g ## are continuous? It seems like I proved the theorem without using this given.
( I know if a function is continuous on a bounded interval the it is Riemann Integrable on that interval, but It seems to me that under the assumption that ## \int_1^{+\infty} f(x)dx ## converges, the assumption that ##f,g ## are continuous is necessary in-order to say that ##g ## is Riemann Integrable on ## [1,\infty) ## , hence I could perform the integrable ##\int_1^{+\infty} g(x)dx ## in the inequality ## \int_R^{+\infty} f(x)dx > \frac{L}{2} \int_R^{+\infty} g(x) dx ## )
2. Was I right that the cases " ## f,g> 0 ## from some place or ##f,g< 0 ## from some place or ## g,f ## have the same sign in ## [1,+\infty) ## " are the all the possible cases?
3. I think that the proofs for the other cases ( besides " ## f,g> 0 ## from some place " ) are very similar for all the cases I have written, is that right?
4. what do you think about the proof? Is it specious? if so, then why? if not, then what could be done better?

The proof can be shortened considerably.

We know, because $f(x)/g(x) \to L > 0$, that there exists $R \geq 1$ such that if $x > R$ then $$\left| \frac{f(x)}{g(x)} - L \right| < \frac L2$$ so that $$\frac L2 < \frac{f(x)}{g(x)} < \frac{3L}2.$$ Continuity of $f$ and $g$ rules out the possibility that they can change sign without passing through zero, and if $f/g$ is to be positive in $(R, \infty)$ then zeros of $f$ and $g$ in $(R, \infty)$ must coincide. But then $f/g$ would be undefined at such a zero, which contradicts the requirement that $f(x)/g(x) \in (\frac{L}2,\frac{3L}2)$. So $f$ and $g$ do not change sign in $(R, \infty)$. It follows that $\int_1^\infty f(x)\,dx$ converges if and only if $\int_R^\infty |f(x)|\,dx$ converges since
• for every $S > R$ we have $\int_R^S |f(x)|\,dx = \pm \int_R^S f(x)\,dx$, and
• $\int_1^\infty f(x)\,dx = \int_1^R f(x)\,dx + \int_R^\infty f(x)\,dx$ converges if and only if $\int_R^\infty f(x)\,dx$ converges
and the same hold for $g$.

Now we have that $$\frac L2 |g(x)| < |f(x)| < \frac {3L}2|g(x)|$$ and integrating from $R$ to $S > R$ yields $$\frac L2\int_R^S |g(x)|\,dx < \int_R^S |f(x)|\,dx < \frac{3L}{2}\int_R^S g(x)\,dx.$$ Taking the limit $S \to \infty$ and applying the integral comparison test to the above shows that
• if $\int_R^\infty |g(x)|\,dx$ converges, then by the second inequality so does $\int_R^\infty |f(x)|\,dx$, and conversely
• if $\int_R^\infty |f(x)|\,dx$ converges, then by the first inequality so does $\int_R^\infty |g(x)|\,dx$
as required.

CGandC
Thank you so much for the time and effort you put into helping me, now I understand what I've wanted. Your proof is much more elegant and understandable.

## 1. What is the limit comparison test for integrals?

The limit comparison test for integrals is a method used to determine the convergence or divergence of an improper integral by comparing it to a known integral with a known convergence or divergence.

## 2. How is the limit comparison test for integrals used?

The limit comparison test for integrals is used by finding a known integral that is similar to the given integral and comparing their convergence or divergence by taking the limit of their quotient.

## 3. What is the condition for the limit comparison test to be applicable?

The condition for the limit comparison test to be applicable is that both integrals must have non-negative integrands and must be evaluated over the same interval.

## 4. What is the significance of the limit comparison test for integrals?

The limit comparison test for integrals is significant because it provides a quick and easy method for determining the convergence or divergence of an improper integral without having to evaluate the integral directly.

## 5. Can the limit comparison test be used to prove convergence or divergence of all integrals?

No, the limit comparison test can only be used to prove the convergence or divergence of integrals with non-negative integrands. It cannot be used for integrals with oscillating or alternating integrands.

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