Week 7 Discussion Notes

Table of Contents

• Direct Comparison

  • Modifying $f\p{x}$
  • Taking Limits

• Sequences

  • Definition
  • Limits

• Homework

  • 8.7.60(a)
  • 8.7.70
  • 8.7.96(c)

Direct Comparison

As a reminder, here's the statement of direct comparison:

Theorem (direct comparison test for integrals)

Suppose $0 \leq g\p{x} \leq f\p{x}$. Then for $-\infty \leq a < b \leq \infty$ (i.e., the upper and lower bounds are allowed to be infinite):

• If $\displaystyle \int_a^b g\p{x} \,\diff{x}$ diverges, then $\displaystyle \int_a^b f\p{x} \,\diff{x}$ diverges as well.
• If $\displaystyle \int_a^b f\p{x} \,\diff{x}$ converges, then $\displaystyle \int_a^b g\p{x} \,\diff{x}$ converges as well.

See the Week 6 Notes for some pictures and intuition for the theorem. So now that we have a good conceptual handle on comparison, how do we use it?

When approaching a problem, you should try to have an idea of whether an integral will converge or diverge (I'll talk about this more during the examples):

• If you want to show that $\int_a^b f\p{x} \,\diff{x}$ converges, you need to find a larger function $g\p{x}$.
• If you want to show that $\int_a^b f\p{x} \,\diff{x}$ diverges, you need to find a smaller (but still non-negative) function $g\p{x}$.

There are two main strategies for getting this inequalities:

Modifying $f\p{x}$

Here, you'll want to modify $f\p{x}$ to get a slightly bigger or slightly smaller function. Usually, $f\p{x}$ is a fraction, so if both the numerator and denominator are non-negative, then:

• To get a bigger function, you can make the numerator bigger or the denominator smaller.
• To get a smaller function, you can make the numerator smaller or the denominator bigger.

Example 1.

(8.7.74)

Determine whether $\displaystyle \int_1^\infty \frac{\diff{x}}{\p{x + x^2}^{1/3}}$ converges or not.

Solution.

First, we need to "feel out" the integral. When $x$ is really big, then $x + x^2 \approx x^2$, since the higher power terms will dominate. So, intuitively, we have

$$\frac{1}{\p{x + x^2}^{1/3}} \approx \frac{1}{\p{x^2}^{1/3}} = \frac{1}{x^{2/3}},$$

which should diverge. So, we need to find a function smaller than the integrand, which can do by making the denominator bigger. The integral is on the interval where $x \geq 1$, so we get $x^2 \geq x$ (by multiplying by $x$ on both sides of the first inequality), which means

$$\p{x + x^2}^{1/3} \leq \p{x^2 + x^2}^{1/3} = \p{2x^2}^{1/3} = 2^{1/3} x^{2/3}.$$

By flipping things over, we get the inequality

$$0 \leq \frac{1}{2^{1/3} x^{2/3}} \leq \frac{1}{\p{x + x^2}^{1/3}}.$$

By $p$-integrals, $\int_1^\infty \frac{\diff{x}}{2^{1/3} x^{2/3}}$ diverges, so by direct comparison, so does $\int_1^\infty \frac{\diff{x}}{\p{x + x^2}^{1/3}}$.

Example 2.

(8.7.75)

Determine whether $\displaystyle \int_0^1 \frac{\diff{x}}{xe^x + x^2}$ converges or not.

Solution.

Like before, we want to get a feel for the integral. When $x$ is really small, $x^2$ doesn't really make a big impact on the integral compared to $x$, so $xe^x + x^2 \approx xe^x$ (similar to how $x + x^2 \approx x^2$ when $x$ is really big). Also, $e^x$ is harmless if $x$ is small, since $e^0 = 1$, so we roughly get

$$\frac{1}{xe^x + x^2} \approx \frac{1}{xe^x} \approx \frac{1}{x},$$

which diverges by $p$-integrals, so we want to show that the integral diverges. We can do this by finding a smaller function, which we can do by making the denominator bigger like before.

Since the integral is over $0 \leq x \leq 1$, we get $x^2 \leq x$ on this interval, which means that $xe^x + x^2 \leq xe^x + x$. Also, $e^x$ is an increasing function, so its maximum is at $x = 1$, which gives $xe^x \leq ex$. Thus, $xe^x + x^2 \leq \p{e + 1}x$ and by flipping things over,

$$0 \leq \frac{1}{\p{e + 1}x} \leq \frac{1}{xe^x + x^2}.$$

The smaller integral diverges by $p$-integrals, so by direct comparison, $\int_0^1 \frac{\diff{x}}{xe^x + x^2}$ diverges, too.

Taking Limits

It might help to brush up on what a limit means. With that said, taking limits will help us prove inequalities for large values of $x$, which is best illustrated through examples:

Example 3.

(8.7.60(a))

Show that if $a > 0$, then $\displaystyle \lim_{x\to\infty} \frac{x^a}{\ln{x}} \implies x^a > 2\ln{x}$ for $x$ large enough.

Solution.

For this problem, we're already given the limit, so we just need to prove the inequality. The definition of the limit tells us that:

$$\text{For all $M > 0$, there exists $N > 0$ such that if $x \geq N$, then $\dfrac{x^a}{\ln{x}} > M$.}$$

Intuitively, the limit tells us that if $x$ is large enough ("... there exists $N > 0$ such that if $x \geq N$ ..."), then $\frac{x^a}{\ln{x}}$ becomes very big as well ("... $\frac{x^a}{\ln{x}} > M$"). The "for all $M > 0$" part tells you that you can make your function as big as you want.

The definition says something works for any $M > 0$, so it means it'll work for a specific $M$. If we set $M = 2$, then there exists $N > 0$ such that whenever $x \geq N$, we have

$$\frac{x^a}{\ln{x}} > 2 \implies x^a > 2\ln{x}.$$

Sequences

Definition

Definition

A sequence $\set{a_n}$ is an ordered list of numbers, where $a_n$ is the $n$-th term of the sequence.

Sequences are just infinitely long lists of numbers. For example:

are both sequences.

Limits

When looking at sequences, we're basically only concerned with limits as $n \to \infty$. Like usual, if the limit exists, then we say that the sequence converges, and if it doesn't, we say that the sequence diverges.

The usual limit laws also apply. For example, assuming all the limits exist,

and so on. One very useful way of taking limits is replacing $n$ with $x$ and calculating them as usual, e.g., with L'Hôpital's rule.

Example 4.

Calculate $\displaystyle \lim_{n\to\infty} \frac{n^2 + 1}{n^2 + 7}$.

Solution.

$$\begin{aligned} \lim_{n\to\infty} \frac{n^2 + 1}{n^2 + 7} &= \lim_{x\to\infty} \frac{x^2 + 1}{x^2 + 7} \\ &\overset{H}{=} \lim_{x\to\infty} \frac{2x}{2x} \\ &= 1. \end{aligned}$$

Warning: If you replace $n$ with $x$ and the limit does not exist, then you can't make any conclusions about the original limit:

Example 5.

Calculate $\displaystyle \lim_{n\to\infty} \sin\p{n\pi}$.

Solution.

You may be tempted to say

$$\lim_{n\to\infty} \sin\p{n\pi} = \lim_{x\to\infty} \sin\p{x\pi} = \mathrm{DNE},$$

since $\sin\p{x\pi}$ oscillates. This is wrong because while $\sin\p{x\pi}$ does indeed oscillate, $\sin\p{n\pi}$ does not:

$$\begin{aligned} \sin\p{0} &= 0 \\ \sin\p{\pi} &= 0 \\ \sin\p{2\pi} &= 0 \\ &\,\,\,\vdots \end{aligned}$$

This means that $\sin\p{n\pi}$ is $0$ for any integer $n$, so

$$\lim_{n\to\infty} \sin\p{n\pi} = \boxed{0}.$$

Homework

8.7.60(a)

See Example 3.

8.7.70

Here, you'll want to recall that $\sinh{x} = \frac{1}{2}\p{e^x - e^{-x}}$, so you can rewrite the integral as

When $x$ is large, $e^{-x} \approx 0$, so the integrand is roughly

so you can tackle this with direct comparison. To get the inequality, see this section or #134 on Campuswire.

8.7.96(c)

From (b), we're given that

So, since the integrand isn't defined at $0$, we need to take a limit to calculate the integral:

If $p < -1$, then $p + 1 < 0$, so so $\p{p + 1}\ln{t}$ is positive as $t \to 0^+$. This means that,

which means that the integral diverges in this case.

If $p > -1$, then $p + 1 > 0$, $\lim_{t\to0^+} t^{p+1}\p{1 - \p{p+1}\ln{t}}$ has the form $0 \cdot \infty$, so you can tackle this using L'Hôpital's.