I'm currently working on some practice exercises in preparation for an exam in a probability theory class and the section on dominated convergence is giving me some trouble. Below are three exercises. I feel like I have the first one right, but then the next two have been driving me crazy. Any help would be much appreciated.
1.) $lim_{n\to\infty} \int_0^1 \frac{1}{\sqrt{t}}e^{-t/n} dt$
Answer: Here, since t is bounded by [0,1], as n goes to infinity, $e^{-t/n}$ converges to 1 (i.e. $e^o$). Thus, we can find dominance in:
$\frac{1}{\sqrt{t}}e^{-t/n} \le \frac{1}{\sqrt{t}}(1)$, or just simply $\frac{1}{\sqrt{t}}$, which is integrable. Thus,
$lim_{n\to\infty} \int_0^1 \frac{1}{\sqrt{t}}e^{-t/n} dt$ = $\int_0^1 lim_{n\to\infty} \frac{1}{\sqrt{t}}e^{-t/n} dt$ = $\int_0^1 \frac{1}{\sqrt{t}}$ = 2
Correct?
2.) $lim_{n\to\infty} \int_0^\infty \frac{r^n}{1+r^{n+2}} dr$
Answer:
Here, finding the dominated integrable function is what I'm getting caught on. It seems easy to say first that clearly $\frac{r^n}{1+r^{n+2}} \le \frac{r^n}{r^{n+2}}$
And we can simplify the right side to $\frac{1}{r^2}$, which is not integrable on our interval..
3.) $lim_{n\to\infty} \int_0^\infty \frac{\sqrt[n]x}{1+x^2} dx$
Answer: Basically the same problem here. First instinct is to say $\frac{\sqrt[n]x}{1+x^2} \le \frac{\sqrt[n]x}{x^2}$
As n goes to infinity, the nth root of any x will converge to 1, so that we can again simplify the right side of the inequality as $\frac{1}{x^2}$, which same as before is not integrable on our interval..
Again, any guidance would be much appreciated.
The first one is correct. For the others, we can decompose the integral to $\int_0^1+ \int_1^{\infty}$, and for each integrand find a dominating function. I will work out the second one (the third one is analogous).
$$\int_0^{\infty}f_n(r)dr = \int_0^1 f_n(r)dr + \int_1^{\infty} f_n(r)dr$$
On $[0,1]$, $f_n(r)\to 0$ almost everywhere (it converges to $0$ except at $r=1$) and:
$$0\le f_n(r)\le \frac1{1+r^2} \in L^1$$
So we get $\lim \int_0^1 f_n(r)dr = 0$.
On $[1,\infty)$, $f_n(r)\to \frac1{r^2}$, and:
$$0\le f_n(r)\le \frac{1}{r^2} \in L^1$$
So $\lim \int_1^{\infty}f_n(r)dr = \int_1^{\infty}\frac1{r^2}dr = 1$
Therefore $\lim \int_0^{\infty}f_n(r)dr = 0+1=1$.