Section 1: Sequences
A sequence is a list of numbers. Below are some examples.
Example: $a_1 = 2, a_2 = 4, a_3 = 6, a_4 = 8, \ldots, a_n = 2n$.
Example: $b_1 = 2, b_2 = 4, b_3 = 8, b_4 = 16, \ldots, b_n = 2^n$.
Example: $f_1 = 1, f_2 = 1, f_3 = 2, f_4 = 3, f_5 = 5, f_6 = 8, f_7 = 13, f_8 = 21, \ldots, f_n = f_{n-1} + f_{n-2}$. This is known as the Fibonacci sequence.
When defining a sequence using a formula based on $n$, $n$ usually starts at 0 or 1, but occasionally at other numbers. A sequence can either converge or diverge. Convergence means that $\lim_{n \to \infty} a_n$ exists. Otherwise, it diverges. This is similar in many ways to limits at infinity of real-valued functions, except that sequences are only defined at integer values. The below example demonstrates this.
Example: Find the limit of $\lim_{n \to \infty} \cos{(2n\pi)}$. Regardless of $n$, the value of $\cos{(2n\pi)}$ is $1$. This means $\lim_{n \to \infty} \cos{(2n\pi)} = 1$. But the real-valued function $\cos{(2x\pi)}$ oscillates between -1 and 1 as $x \to \infty$.
A sequence can diverge in many ways. For example $a_n = n$ diverges to $\infty$, $b_n = -\sqrt{n}$ diverges to $-\infty$, and $\cos{n}$ diverges in no special way (it oscillates between -1 and 1 with no limit).
When thinking about convergence or divergence, where the sequence starts does not matter. For example, $a_n = \dfrac{1}{n^2}$ converges to 0 whether we start at $a_1 = 1$ or at $a_8 = 1/64$.
The usual limit laws of real-valued functions apply to sequences as well. So does the Squeeze Theorem.
Example: Let $a_n = \dfrac{\sin{n}}{7^n}$. Let's find the limit of $a_n$. Since $-1 \leq \sin{n} \leq 1$, we can write the inequality $\dfrac{-1}{7^n} \leq a_n \leq \dfrac{1}{7^n}$. Since both $\dfrac{-1}{7^n}$ and $\dfrac{1}{7^n}$ have a limit of 0, this means that $0 \leq \lim_{n \to \infty} a_n \leq 0$, so $\lim_{n \to \infty} a_n = 0$.
Mini-Quiz: Find $\lim_{n \to \infty} \left(\dfrac{1}{2}\right)^n\cos\left({\dfrac{1}{n}}\right)$.
It is possible to use L'Hospital's Rule, but you need to be careful. In order to use L'Hospital's Rule on a sequence of the form $\dfrac{f(n)}{g(n)}$, you need to consider the real-valued function $\dfrac{f(x)}{g(x)}$ first. It makes no sense to differentiate a sequence which is only defined for integers. If $\dfrac{f(x)}{g(x)}$ has a limit $L$ as $x \to \infty$, then $\lim_{n \to \infty} \dfrac{f(n)}{g(n)} = L$ also. But the converse is not true, as we saw in the example above with $\cos{(2n\pi)}$.
Now let's see an example of how to apply L'Hospital's Rule.
Example: Let $a_n = n^{1/n}$. This has the form "$\infty^0$", which is an indeterminate form. However, we require either $0/0$ or $\infty/\infty$ to use L'Hospital's Rule. So let $b_n = \ln{a_n} = \dfrac{\ln{n}}{n}$. This takes the form $\infty/\infty$. So let $b(x) = \dfrac{\ln{x}}{x}$. Take the derivative of top and bottom to get $\dfrac{1/x}{1} \to 0$ as $x \to \infty$. This means $\lim_{n \to \infty} b_n = 0$. Since $b_n = \ln{a_n}$, we know that $a_n = e^{b_n}$. Thus, $\lim_{n \to \infty} a_n = e^0 = 1$.
Mini-Quiz: Find $\lim_{n \to \infty} \left(1 - \frac{2}{n}\right)^n$.
Perhaps it's useful to have an ordering of common functions, just to develop some intuition. The ordering is as follows:
$$(\log{n})^k \ll n^k \ll a^n \ll n! \ll n^n.$$ What this means is that polynomials grow faster than powers of logarithms, exponentials grow faster than polynomials, factorials grow faster than exponentials, and hypergeometric sequences grow faster than factorials. This means, for example, that $$\lim_{n \to \infty} \frac{\sqrt{n}}{(\log{n})^3} = \infty, \lim_{n \to \infty} \frac{2^n}{n!} = 0,$$ etc. When you see a problem, you should look at the associated real-valued function and use L'Hospital's Rule (but other strategies need to be used to analyze factorials). But your intuition should reflect this ordering.
If you haven't seen factorials before, $n!$ just means the first $n$ positive integers multiplied together. So $5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5$, and so on. Sometimes $0!$ is defined to be 1. One useful fact when manipulating factorials is that $(n+1)! = n!(n+1)$.
The Monotone Convergence Theorem says that if a sequence $a_n$ is either increasing or decreasing, and is bounded, then $a_n$ converges.
Example: Let $a_1 = 3, a_n = 1 + \dfrac{a_{n-1}}{2}$. Let's try to find the limit of $a_n$. Let's write out a few terms to get a sense of what's going on: $a_1 = 3, a_2 = 2.5, a_3 = 2.25, a_4 = 2.125$. It certainly appears that $a_n$ is decreasing. One way to prove this is to let $f(x) = 1 + x/2$. You can see that $f'(x) = 1/2 > 0$, so $f(x)$ is an increasing function. Remember that this means that $f(x) \geq f(y)$ whenever $x \geq y$. We know that $a_1 \geq a_2$. So $f(a_1) = a_2 \geq f(a_2) = a_3$. Similarly, since $a_2 \geq a_3$, $f(a_2) = a_3 \geq f(a_3) = a_4$, and so on. So $a_1 \geq a_2 \geq \ldots \geq a_n \geq \ldots$, which means $a_n$ is decreasing.
To show the sequence is bounded: it is decreasing, so an upper bound is the first term: 3. The tricky part is finding the lower bound. It appears that 2 is a lower bound. So let's try to prove this. The goal is to show $a_n = 1 + \dfrac{a_{n-1}}{2} \geq 2$. Subtract off the 1, and this is true as long as $\dfrac{a_{n-1}}{2} \geq 1$. Multiply by 2, and this is true as long as $a_{n-1} \geq 2$. Therefore, $a_n \geq 2$ as long as $a_{n-1} \geq 2$. This logic works all the way back as long as $a_1 \geq 2$. But $a_1 = 3$, which is certainly bigger than 2. (In other words $a_{n-1} \geq 2$ implies $a_n \geq 2$, and $a_1 \geq 2$, so $a_2 \geq 2$, which means $a_3 \geq 2$, etc.) So $a_n$ is bounded.
The Monotone Convergence Theorem says that $a_n$ converges. But to what value? Whenever a sequence $a_n$ converges, sequences like $a_{n+1}, a_{n+5}, a_{2n}$ all converge to the same limit. So let's call the limit $L$. From the equation $a_{n} = 1 + \dfrac{a_{n-1}}{2}$, we know that $L = 1 + L/2$, or $L/2 = 1$. This means $L = 2$. This problem was quite involved, but know that it's one of the most difficult questions you'll see on sequences.
Mini-Quiz: Let $b_1 = \sqrt{5}, b_n = \sqrt{5 + b_{n-1}}$. Find $\lim_{n \to \infty} b_n$.