Sequence and series

A sequence is a function defined on the set of positive integers. This means that if you give me a positive integer $5$, I should be able to give you some object $a_{5}$, labeled with the number you gave me, which is $5$ in this case.

For example, we can think about the "identity sequence" given by $a_{n} = n$. That is, this sequence gives you back the number your give it as an input. Another example can be $a_{n} = n^{2}$. Sometimes, you may not be able to write down the formula for $a_{n}$ explicitly. For instance, define $a_{n}$ to be the $n$-th digit of $\pi = 3.141592 \dots$ after the dot. We have $a_{1} = 1, a_{2} = 4, a_{3} = 1, a_{4} = 5, \dots.$ However, because $\pi$ is an irrational number, so there is no way you can write down a general formula for $a_{n},$ which we call the $n$-th term of the sequence $(a_{i})_{i \geq 1}.$

Remark. Note that most of the time, it is perfectly fine to extend the domain of the sequence from positive integersA sequence is a function defined on the set of positive integers. This means that if you give me a positive integer $5$, I should be able to give you some object $a_{5}$, labeled with the number you gave me, which is $5$ in this case.

For example, we can think about the "identity sequence" given by $a_{n} = n$. That is, this sequence gives you back the number your give it as an input. Another example can be $a_{n} = n^{2}$. Sometimes, you may not be able to write down the formula for $a_{n}$ explicitly. For instance, define $a_{n}$ to be the $n$-th digit of $\pi = 3.141592 \dots$ after the dot. We have $a_{1} = 1, a_{2} = 4, a_{3} = 1, a_{4} = 5, \dots.$ However, because $\pi$ is an irrational number, so there is no way you can write down a general formula for $a_{n},$ which we call the $n$-th term of the sequence $(a_{i})_{i \geq 1}.$

Remark. Note that most of the time, it is perfectly fine to extend the domain of the sequence from positive integers to non-negative integers. For example, the sequence given by the formula $a_{n} = n$ can be thought over all the non-negative integers or even negative integers. It is of course okay to extend this formula for all real numbers, but we usually don't call it a sequence if it is over all real numbers. There is a reason for this, but discussing this digresses too much from our main discussion, so we won't discuss this.

We can take the limit of sequence, when ever the limit exists. For example, we have

\[\lim_{n \rightarrow \infty}\frac{1}{n} = 0.\]

One important property of real numbers is the following. The name of this property is "completeness", although your book (p.477, Theorem 9.1) does not mention this terminology.

Completeness proeprty. Any increasing sequence bounded above is convergent.

Exercise. Determine whether the sequence given by the following formula is increasing:

\[a_{n} = \left(1 + \frac{1}{n}\right)^{n}.\]

For the problem above, you might find the following hint useful:

Consider $f(x) = (1 + 1/x)^{x},$ for $x > 0.$ Let $g(x) := \ln(f(x)) = x \ln(1 + 1/x)$ so that $$g'(x) = \ln\left(1 + \frac{1}{x}\right) + \frac{x}{1 + 1/x}\left(-\frac{1}{x^{2}}\right) = \ln\left(1 + \frac{1}{x}\right) - \frac{1}{x + 1}.$$

Hence $g'(x) > 0$ for $x \geq 1.$ This shows that $g(x) = \ln(f(x))$ is increasing for $x \geq 1,$ so $f(x)$ is increasing for $x \geq 1.$ This implies that $f(n) = a_{n}$ is increasing for $n = 1, 2, 3, \dots.$

Remark. How do we know that $g'(x) > 0$ for $x \geq 1$? First, note that $g'(1) = \ln(2) - 1/2 > 0$ because $2 > e^{1/2}$, because $4 > e.$ Now, take the derivative of $g'(x)$ to have $$(g'(x))' = g''(x) = \frac{1}{1 + 1/x}\left(- \frac{1}{x^{2}}\right) + \frac{1}{(x + 1)^{2}} = -\frac{1}{x(x + 1)} + \frac{1}{(x+1)^{2}} > 0,$$ for $x \geq 1.$ Thus, we must have $g'(x) \geq \ln(2) - 1/2 > 0$ for all $x \geq 1.$

We have checked that the sequence $a_{n} = (1 + 1/n)^{n}$ is increasing. The sequence is also bounded above by $3$. Checking this is same as checking $1 + 1/n \leq 3^{1/n},$ which can be checked by checking $1 + x \leq 3^{x}$ for $0 \leq x \leq 1,$ which can be checked by differentiation trick we have introduced above. Therefore, by the completeness property, the limit of $a_{n}$ exists, and we call this limit $e.$ That is, we define $$e := \lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^{n}.$$

A geometric sequence is a sequence such that that each pair of the consecutive terms have the same ratio. If such a common ratio $r$ is non-zero, then we can consider this sequence to be given by $a_{n+1}/a_{n} = r$ for all $n \geq 1.$ More explicitly, you may deduce that

\[a_{n} = a_{1}r^{n-1}.\] 

A series is a sum of the terms of a sequence. You can have finite series or infinite series. There is a version of $p$-test in the series world. 

$p$-series. Given any real number $p,$ the series

\[\sum_{n=1}^{\infty}\frac{1}{n^{p}}\]

converges if and only if $p > 1.$ We refer to this as "$p$-series", although this is not a standard math terminology outside Math 116.

Question. Is there a more general phenomenon?

Answer. Yes! This is what we call the integral test. Please read the link, and at least get comfortable with the statement. Alternatively, you may read p.492 of your book.

Exercise. Solve #4 of Exam 3, Fall 2016.

Optional reading. You may consider $p$-series as a function in $p$, and it turned out that this is one of the most important functions in mathematics, called "zeta function". I like thinking about it, since it often breaks data about integers into atomic pieces about prime numbers.

A geometric series is a sum of a geometric sequence. Say $a_{1}, a_{2}, a_{3}, \dots$ is the geometric sequence of ratio $r$. Then we may consider their partial sums $$s_{n} := a_{1} + \cdots + a_{n}.$$

Exercise. Show that $s_{n} - s_{n-1} = a_{n}$ for $n \geq 1.$ Note that this holds for any sequence, not just for geometric sequences. This proves that if $s_{n}$ converges, then $a_{n}$ converges to $0$, which is called the general term test. (You might have to call it "$n$-th term test" for your exam, but this terminology is misleading.) A more useful way of thinking about this test is the following.

Term test for divergence. If $a_{n}$ does not go to $0,$ then $\sum_{n=1}^{\infty}a_{n}$ diverges.

Exercise. Using the fact that $a_{n} = a_{1}r^{n-1}$, show that if $r \neq 1,$ then

\[s_{n} = \frac{a_{1}(r^{n} - 1)}{r - 1} = \frac{a_{1}(1 - r^{n})}{1 - r}.\]

In particular, if $-1 < r < 1,$ then

 $$a_{1} + a_{2} + a_{3} + \cdots = \lim_{n \rightarrow \infty}s_{n} = \frac{a_{1}}{r - 1},$$ and you must remember this fact about geometric series!

Exercise. Show that $0.99999 \dots = 1.$

Probability

Introduction to probability distributions. The discussion here might be too advanced for Math 116 students, so I highly recommend you read through Section 8.7 and 8.8 of your book. Nevertheless, I think the book needs a bit more explanations, so I am just writing some here. Don't worry. I won't give proofs in class!

A continuous real-valued function $p(x)$ is called a probability density function (pdf) if the following two conditions are satisfied:

1. $p(x) \geq 0$ for all $x.$
2. $\int_{-\infty}^{\infty}p(x)dx = 1.$

Thus, given such a function $p(x),$ you can model some notion of probability. For example, your model can design (if chosen at random) "the probability that a person's height $x$ is between $5$ feet and $6$ feet" as the quantity $$\int_{x=5}^{x=6} p(x) dx.$$ Of course, you might want to pick $p(x)$ well enough that your probability is not too far from reality. (This is very nontrivial in practice.)

The cumulative distribution function (cdf) of a pdf $p(x)$ is $$P(t) := \int_{x= - \infty}^{x=t}p(x)dx.$$

Exercise. If $P(t)$ is the cdf of a pdf $p(x),$ then show that $P'(t) = p(t).$

Exercise. If $P(t)$ is the cdf of a pdf $p(x),$ then show that

1. $P(t)$ is non-decreasing,
2. $\lim_{t \rightarrow -\infty}P(t) = 0$,
3. $\lim_{t \rightarrow \infty}P(t) = 1$, and
4. $\int_{a}^{b}p(x)dx = P(b) - P(a).$

Exercise. Book 8.7: #1, #5, #14, #24.

Given any p.d.f. $p,$ its mean is $$\mu(p) := \int_{-\infty}^{\infty}x p(x) dx,$$ and we will only consider the case when this is a convergent integral.

Remark. For this, we need to ensure that $p(x)$ decays fast enough. For instance, if we let $c = \int_{1}^{\infty}x^{-2} dx$ and define $p(x) = c^{-1}x^{-2}$ for $x \geq 1$ and $0$ elsewhere, then $p(x)$ is a legitimate p.d.f, while we have $$\mu(p) = c^{-1}\int_{1}^{\infty} \frac{1}{x} dx = \infty,$$ which we cannot use in practice.

The standard deviation of $p$ is defined by $$\sigma(p) := \left(\int_{-\infty}^{\infty}(x - \mu(p))^{2}f(x)dx\right)^{1/2}.$$ Again we need to assume that $p(x)$ decays quite fast as it goes away from $0$ to $\infty$ or $-\infty.$

Theorem 1. If $p$ is a p.d.f that decays fast enough, we have
$$\sigma(p)^{2} = \int_{-\infty}^{\infty}x^{2}p(x)dx - \mu(p)^{2}.$$ Proof. For convenience, write $\mu = \mu(p).$ By definitions of the standard deviation and the mean of $p$, we have
\begin{align*} \sigma(p)^{2} &= \int_{-\infty}^{\infty}(x - \mu(p))^{2}p(x)dx \\ &= \int_{-\infty}^{\infty}(x^{2} - 2\mu x + \mu^{2})p(x)dx \\ &= \int_{-\infty}^{\infty}x^{2}p(x) dx - 2\mu\int_{-\infty}^{\infty}xp(x)dx + \mu^{2}\int_{-\infty}^{\infty}p(x)dx \\ &= \int_{-\infty}^{\infty}x^{2}p(x) dx - 2\mu\mu + \mu^{2} \cdot 1 \\ &= \int_{-\infty}^{\infty}x^{2}p(x) dx - \mu^{2}. \end{align*}
Note that for the second to the last equality, we have used the condition $$\int_{-\infty}^{\infty}p(x)dx = 1,$$ which follows from the assumption that $p$ is a p.d.f. This finishes the proof.

Q.E.D.

We will compute the mean and the standard deviation of the following probability distribution function (p.d.f.): $$f_{m,s}(x) = \frac{e^{-(x - m)^{2}/(2s^{2})}}{s\sqrt{2 \pi}},$$ where $m, s$ are constants and $s > 0.$ It is evident that $f_{m,s}(x) \geq 0$ for all $x$ because the exponentiation always gives us a positive value. However, to show $f_{m,s}(x)$ is a p.d.f., we also need to check that $$\int_{-\infty}^{\infty}f_{m,s}(x)dx = 1,$$ and this is not so obvious.

How to check this in your future. The easiest case is when $s = 1/\sqrt{2}$ and $m = 0$ (which can be done in Math 215 using the change of variables in polar coordinates). Then the substitution method to manipulate the integral will do the job.

Theorem 2. We have $\mu(f_{m,s}) = m.$

Proof. By definition of the mean, we have
\[\mu(f_{m,s}) = \int_{-\infty}^{\infty}xf_{m,s}(x) dx = \int_{-\infty}^{\infty} \frac{xe^{-(x - m)^{2}/(2s^{2})}}{s\sqrt{2\pi}}dx.\]
Consider the substitution
\[u = \frac{x - m}{s\sqrt{2}},\]
which comes with
\[du = \frac{dx}{s\sqrt{2}}.\]
Moreover, note that $x = s\sqrt{2} \cdot u + m.$ Hence, following the computation we have made above, we get
\begin{align*}\mu(f_{m,s}) &= \frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}(s\sqrt{2} \cdot u + m)e^{-u^{2}}du \\ &= \frac{m}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^{2}}du \\ &= m.\end{align*}
Note that the second to the last equality uses that
\[\int_{-\infty}^{\infty}ue^{-u^{2}}du = 0,\]
which is a property for any odd function. The last equality uses the fact that $f_{0,1/\sqrt{2}}$ is a p.d.f.

Q.E.D.

The following turned out to be more difficult that I thought:

Theorem 3. We have $\sigma(f_{m,s}) = s.$

Proof. By definition of the standard deviation and using Theorem 2 which says $\mu(f_{m,s}) = m$, we have
\[\sigma(f_{m,s})^{2} = \int_{-\infty}^{\infty}(x - m)^{2}f_{m,s}(x)dx = \int_{-\infty}^{\infty} \frac{(x - m)^{2}e^{-(x - m)^{2}/(2s^{2})}}{s\sqrt{2\pi}}dx.\]
Consider the substitution
\[t = \frac{x - m}{s\sqrt{2}},\]
which comes with
\[dt = \frac{dx}{s\sqrt{2}}.\]
Moreover, note that $x - m = st\sqrt{2}.$ Hence, following the computation we have made above, we get
\[\sigma(f_{m,s})^{2} = \frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}2s^{2}t^{2}e^{-t^{2}}dt = \frac{4s^{2}}{\sqrt{\pi}}\int_{0}^{\infty}t^{2}e^{-t^{2}}dt,\]
where the last integral uses a property of even functions. Hence, we can just focus on computing the last integral. Since we only need to think about $t \geq 0,$ we can use the substitution
\[u = t^{2},\]
which comes with $du = 2tdt.$ Continuing the computation above, we have
\begin{align*}\sigma(f_{m,s})^{2} &= \frac{4s^{2}}{\sqrt{\pi}}\int_{0}^{\infty}t^{2}e^{-t^{2}}dt \\ &= \frac{2s^{2}}{\sqrt{\pi}}\int_{0}^{\infty}u^{1/2}e^{-u}du \\ &= \frac{2s^{2}}{\sqrt{\pi}}\Gamma(3/2), \end{align*}

where

\[\Gamma(z) = \int_{0}^{\infty}u^{z-1}e^{-u}du.\]

This is called the Gamma function, which you can read about here. One of the properties in the link says that we have the duplication formula:

\[\Gamma(z)\Gamma(z + 1/2) = 2^{1 - 2z} \sqrt{\pi}\Gamma(2z).\]

It is also not difficult to figure out $\Gamma(n) = (n-1)!$ for all integer $n \geq 1.$ Hence taking $z = 1$ in the duplication formula, we get $$\Gamma(3/2) = 2^{-1}\sqrt{\pi}\Gamma(2) = 2^{-1}\sqrt{\pi}.$$ Combining with the previous computation, we have $$\sigma(f_{m,s})^{2} = \frac{2s^{2}}{\sqrt{\pi}}\Gamma(3/2) = \frac{2s^{2}}{\sqrt{\pi}}2^{-1}\sqrt{\pi} = s^{2}.$$

This implies that $\sigma(f_{m,s}) = s,$ as desired.

Q.E.D.

Upshot: The distribution given by the p.d.f $f_{m,s}$ is called the normal distribution with mean $m$ and standard deviation $s.$

Median. Given a pdf $p(x),$ the median of the distribution given by $p(x)$ is the value $T$ such that $$\int_{-\infty}^{T}p(x) dx  = 1/2.$$ Roughly, the median estimates what's in the middle, while the mean means the average.

Exercise. Suppose that $p(x)$ is a pdf that is symmetric to the vertical line $x = a,$ where $a$ is a fixed real number.

1. What is the median given by the distribution?
2. What is the mean given by the distribution?

Work

In this posting, we study the concept of work. You may find more detailed exposition with various examples in Section 8.5 of your book.

Slogan: Work is "energy".

Linear fashion (local/micro work). Work = Force $\times$ Distance

Nonlinear fashion (global/total work). When $x$ is a position that moves from $a$ to $b$ and $F(x)$ is the force acting at $x$, then

\[\text{Work} = \int_{a}^{b}F(x) dx.\]

Gravitational acceleration from physics. It is quite a remarkable observation that on Earth, if you drop any two objects from the same height when there is no air at presence, both objects will hit the ground at the same time. This means that any object has a constant acceleration, not depending on the object, when we do not count air resistance. In the unit (meter/second)/second = $m/s^{2}$, this constant is denoted as $g$, and it is known that it is approximately equal to $9.8$, so when the problem does not give you any information about gravitation, you should start by write something like "Let $g$ be the gravitational acceleration in $m/s^{2}$" or "Let $g = 9.8 m/s^{2}$ be the estimated gravitational constant."

Force is mass times acceleration when the quantities are constant. In classical mechanics, it is assumed that the force is equal to mass times acceleration (a.k.a., the "law" $F = ma$). This is something we assume in this course as well, if the mass and acceleration are assumed to be constant. However, this will rarely happen. This formula will usually make sense when we compute micro-force.

Exercise. Suppose that the mass of your book is $2$kg. If you have lifted your book $1.5$ meters off the floor, then how much work have you done?

Exercise. Say you are pushing a ball whose mass is $1$kg along the graph $y = x$ from point $(0, 0)$ to $(1, 1)$. When you are done, how much work would you have done? (Suppose that the unit for distance on our $xy$-plane is in meters.)

Exercise (Hard). Say you are pushing a ball whose mass is $1$kg along the graph $y = x^{2}$ from point $(0, 0)$ to $(1, 1)$. When you are done, how much work would you have done? (Suppose that the unit for distance on our $xy$-plane is in meters.)

(Hint: This is a very hard exercise for any student who is taking Math 116 from my past experience. Writing $g$ for the constant gravitational acceleration, if you compute correctly the instantaneous force for the particle with $x$-coordinate $x$ is equal to $g \sin(\theta(x))$ and the instantaneous distance would be $\Delta x / \cos(\theta(x))$. Here, note that $\theta(x)$ is the angle between the tangent line of the point $(x, y)$ on the graph $y = x^{2}$ and the $x$-axis. Note that by taking derivative, you see $\tan(\theta(x)) = 2x,$ and this would let you compute the integral.)

Remark. The unit for the work in the last exercise is in joules (i.e., $kg \cdot m^{2}/s^{2}$, which is often denoted as $J$).

Exercise. Solve 3a and 3b of Exam 1, Winter 2017.

Exercise. Solve 9a, b, and c of Exam 1, Winter 2017.

Exercise. Solve 9 of Exam 1, Winter 2016.

Exercise. Solve 9 of Exam 1, Fall 2015.

Parametrization and arc length

Parametrization: case study. Consider our favorite example $y = x^{2}$. We may consider the curve from $(0, 0)$ to $(1, 1)$. Naturally, you think about a particle moving along the curve. We may consider the particle to be $(x(t), y(t)) = (t, t^{2})$ because then it will satisfy $y(t) = x(t)^{2}$. Notice that $$\frac{y'(t)}{x'(t)} = 2x,$$ which is equal to $y'(x) = 2x.$

In general, when $y$ is a differentialble function in $x$ and $x$ is a differentiable function in $t$ so that

• $y = y(x)$;
• $x = x(t)$.

Then we have $y(t) = y(x(t))$, so $y'(t) = y'(x(t))x'(t)$ by Chain Rule. Hence, we get $$y'(x) = y'(x(t)) = \frac{y'(t)}{x'(t)},$$ as long as $x'(t) \neq 0.$ In the above example, we had $x(t) = t$ and $y(x) = x^{2}.$ This formula can be remembered as: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$ This gives a nice way to take derivatives.

Example. Consider the implicit function $y(x)$ that is given by the unit circle $x^{2} + y^{2} = 1.$ We can locally compute $y'(x)$ as follows: first, consider the parametrization given by

• $x(t) = \cos(t)$ and
• $y(t) = \sin(t)$.

Then $y'(\cos(t)) = y'(x(t)) = x'(t)/y'(t) = -\sin(t)/\cos(t) = -y(t)/x(t),$ so you can see that $y'(x) = -y/x$ when $x \neq 0$.

Arc length: case study. You know the length from $(0,0)$ to $(1,1)$ along the line $y = x$. Can you compute the length from $(0,0)$ to $(1,1)$ along the curve $y = x^{2}$?

Write $f(x) = x^{2}$. To compute the length, let's break the interval $[0, 1]$ into pieces. Each subinterval would have length $\Delta x$ and you can estimate the length you want by summing up lengths of line segments $\sqrt{(\Delta x)^{2} +(f(x + \Delta x) - f(x))^{2}}.$ Note that $$\sqrt{(\Delta x)^{2} +(f(x + \Delta x) - f(x))^{2}} = \sqrt{1 + \left(\dfrac{f(x + \Delta x) - f(x)}{\Delta x}\right)^{2}}\Delta x,$$ so when you sum them up and take $\Delta x \rightarrow 0$, you should get $$\textbf{Arc length} = \int_{0}^{1}\sqrt{1 + f'(x)^{2}}dx = \int_{0}^{1}\sqrt{1 + 4x^{2}} dx.$$

In general, if you have a smooth function $y = f(x)$, then $$\textbf{Arc length from } x = a \textbf{ to }  x = b \text{ is } \int_{a}^{b}\sqrt{1 + f'(x)^{2}}dx.$$

If you have a parametrization $(x(t), y(t)),$ again, thinking $y(t) = y(x(t))$, you may apply the change of variable in the above formula. That is, you consider $x = x(t)$ so that $dx = x'(t)dt$. Find $t = c$ such that $x(c) = a$ and $t = d$ such that $x(d) = b$, you get

\[\int_{c}^{d}\sqrt{1 + (y'(t)/x'(t))^{2}}x'(t)dt = \int_{c}^{d}\sqrt{x'(t)^{2} + y'(t)^{2}}dt.\]

You can consider the above as the arc length of the particle $(x(t), y(t))$ from time $t = c$ to $t = d$.

Memorization trick. Given a particle $(x(t), y(t))$, its velocity is $(x'(t), y'(t))$ and its speed is the size of the velocity, which is $\sqrt{x'(t)^{2} + y'(t)^{2}}$. You notice that:

Arc length is the integral of speed!

Exercise. Using arc length of the parametrization $(2\cos(t), 2\sin(t))$ of the circle of radius $2$ for $0 \leq t \leq 2\pi$, derive the usual fact that its circumference is equal to $4\pi$.

Exercise. Solve #4b of Exam 1, Winter 2017.

Exercise. Solve #6c of Exam 1, Fall 2016.

Exercise. Solve #3 of Exam 1, Fall 2015.

Exercise. Solve #8 of Exam 1, Fall 2015.

Area, volume, and mass

It is somewhat impossible to explain this concept without drawing pictures, so I am just going to let your book take care of the conceptual explanations.

Exercise. Read Sections 8.1, 8.2, and 8.4.

Instead, we will work on problems in class.

Improper integral (with singularity)

Recall from the last time that the integral $$\int_{1}^{\infty}\frac{1}{x^{p}}dx$$ is finite exactly for $p > 1.$ To turn it around, we may ask the same question for the following integral: $$\int_{0}^{1} \frac{1}{x^{p}} dx.$$ First, we need to know what we are talking about, namely, we have the problematic point $x = 0,$ because we cannot put that on the denominator of the integrand. (Such a point is called a "singularity".) What do we do? We just avoid it slightly and approach the singularity instead: we define $$\int_{0}^{1} \frac{1}{x^{p}} dx = \lim_{\epsilon \rightarrow 0+} \int_{\epsilon}^{1} \frac{1}{x^{p}} dx.$$ Using the 1st fundamental theorem of calculus, you will realize that the integral is finite precisely when $p < 1.$

Exercise. Do all the problems in this problem set!

Improper integrals (with point at infinity)

Motivation. We will eventually talk about infinite sums in this course, and they are quite tricky. For example, consider $$1 + \frac{1}{2} + \frac{1}{3} + \cdots.$$ It looks like each time we add something smaller than the previous term, but in fact, we have $$1 + \frac{1}{2} + \frac{1}{3} + \cdots = \infty.$$ How do we know? First, do the following exercise:

Exercise. By drawing a good picture, explain that $$1 + \frac{1}{2} + \frac{1}{3} + \cdots \geq \int_{1}^{\infty}\frac{1}{x}dx.$$ (Hint: the numbers we add will be the areas of certain boxes.)

Now, we have $$\begin{align*} \int_{1}^{\infty}\frac{1}{x}dx &= \lim_{r \rightarrow \infty}\int_{1}^{r}\frac{1}{x}dx \\ &= \lim_{r \rightarrow \infty}\left.\ln(|x|)\right|_{1}^{r} \\&= \lim_{r \rightarrow \infty}\ln(r) \\&= \infty. \end{align*}$$ This lets us see that the infinite sum is infinite.

Exercise. Determine whether the following sum is finite or infinite: $$1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \cdots.$$ This will be a part of somewhat general behavior:

Exercise. Figure out exactly which real number $p$ makes the following integral finite: $$\int_{1}^{\infty}\frac{1}{x^{p}}dx.$$ The finite integrals will be called converegent, while the infinite integrals will be called divergent.

Exercise. Read Section 7.6.

Exercise. Determine whether the following integral is convergent or divergent: $$\int_{1}^{\infty} \frac{1}{(x+4)^{2}}dx.$$ (Hint: Use substitution trick.)

Exercise. Determine whether the following integral is convergent or divergent: $$\int_{1}^{\infty} \frac{x}{(x+4)^{2}}dx.$$ (Hint: Use substitution trick and then split the integral into two.)

Exponential function. One improper integral involving $\infty$ that comes up a lot is the exponential decay. Namely, the function $e^{-x}$ decays so fast that $$\int_{1}^{\infty} e^{-x}dx$$ is finite.

Exercise. Explain why $$\frac{1}{x^{2}} \leq e^{-x}$$ for large enough $x.$ Using this information, show that the following integral is finite: $$\int_{1}^{\infty} e^{-x}dx.$$ (Hint: Try to think about using L'H on $x^{2}/e^{x}.$)

Exercise. Compute $$\int_{1}^{\infty} e^{-x}dx$$ as an exact value.

Exercise. Figure out which $a$ makes the following integral finite: $$\int_{1}^{\infty} e^{ax}dx.$$ (Hint: For $a \geq 0,$ note that $e^{ax} \geq 1$ for all $x.$)

Exercise. Determine the convergence of the following integral: $$\int_{3}^{\infty} x^{3}e^{-x}dx.$$ (Hint: Recall that $e^{x} \geq x^{5}$ for large enough $x.$ Why does this help?)

Exercise. Determine the convergence of the following integral: $$\int_{3}^{\infty} \ln(x)(e^{-x} + x^{-1})dx.$$ (Hint: Try to split the integral.)

Exercise. Determine the convergence of the following integral: $$\int_{3}^{\infty} x^{-3000}e^{x}dx.$$ (Hint: $e^{x} \geq x^{3000}$ eventually in $x$.)

Exercise. Determine the convergence of the following integral: $$\int_{3}^{\infty} \sin^{2}(x)x^{-2}dx.$$ (Hint: $-1 \leq \sin(x) \leq 1$.)