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# Week 4 Wednesday 7/12 brief notes.
## A warm-up for last time.
**Example.** Determine if the improper integral $\displaystyle\int_{4}^{\infty} \frac{\ln(x)}{x}$ converge or diverge.
**Example.** Determine if the improper integral $\displaystyle\int_{4}^{\infty} \frac{\ln(x)}{x^{2}}dx$ converge or diverge.
**Example.** Determine if the improper integral $\displaystyle\int_{4}^{\infty} \frac{\ln(x)}{x^{3}}dx$ converge or diverge.
## Sequences.
A **sequence** is a list of number in a fixed **order** $$
(a_{1}, a_{2},a_{3},a_{4},\ldots)
$$where each $a_{k}$ is some number, and $k$ is the index. We call $a_{k}$ the $k$-th term of this sequence. Sometimes we denote $(a_{1},a_{2},\ldots)$ by the short hand $(a_{n})$ or $(a_{n})_{n=1}^{\infty}$. If we do not state what the starting index is, assume it is $1$ or $0$, or anything that is sensible. We are going to be only interested in sequences with infinitely many terms.
**Example.**
This sequence $\displaystyle(\frac{1}{2^{n}})_{n=1}^{\infty}$ looks like this $$
\left( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},\ldots \right)
$$ and it keeps going. The rule of the $n$th term of the sequence is given by $\displaystyle a_{n}=\frac{1}{2^{n}}$.
Quite often, if we are given **explicitly** the formula of the $n$th term $a_{n}=f(n)$, then we know directly what each $a_{n}$ is using the formula. For instance if $a_{n}=\frac{3n}{2n+1}$, then $a_{10}=\frac{30}{21}$.
Sometimes, if one is given only a first few terms of a sequence, one could **try** to guess an explicit formula for it, but this guess is never a unique answer. That is, if we are given only finitely many terms in the sequence, there are infinitely many different formulas that can give those terms! Nevertheless, one can still try to find one that fit.
**Example.** Come up with an explicit rule $a_{n}$if we have the following $$
\sqrt{4}, \sqrt{5}, \sqrt{6}, \sqrt{7},\ldots
$$
One answer could be $a_{n}=\sqrt{n}$ and we start with $n=4$.
Another answer could be $a_{n}=\sqrt{n+3}$ and we start with $n=1$.
Another answer could be $a_{n}=\sqrt{5 - (19 n)/6 + (35 n^2)/12 - (5 n^3)/6 + n^3/12}$, where we start with $n=1$ !!! Check it out! Try putting in $n=1,2,3,4,...$ you will get the first four term to be $\sqrt{4}, \sqrt{5}, \sqrt{6}, \sqrt{7}$. But the next terms will be $\sqrt{10},\sqrt{19},\sqrt{40}, \sqrt{81},...$, we weren't given what the next few terms should be so anything could be there! (In any case those guessing the pattern IQ questions are all kind of silly.)
Sometimes the sequence can be given **recursively**. Meaning we are given some initial values, and then a rule relating the terms together:
**Example.** Fibonacci sequence.
Let $a_{1}=$ and $a_{2}=1$, and for $n\ge 3$ we have $a_{n}=a_{n-1}+a_{n-2}$.