Prove limit comparison test for Integrals

[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?

 

[pasmith]

The proof can be shortened considerably.

We know, because f(x)/g(x) \to L &gt; 0, that there exists R \geq 1 such that if x &gt; R then <br /> \left| \frac{f(x)}{g(x)} - L \right| &lt; \frac L2 so that <br /> \frac L2 &lt; \frac{f(x)}{g(x)} &lt; \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 &gt; 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 <br /> \frac L2 |g(x)| &lt; |f(x)| &lt; \frac {3L}2|g(x)| and integrating from R to S &gt; R yields <br /> \frac L2\int_R^S |g(x)|\,dx &lt; \int_R^S |f(x)|\,dx &lt; \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.