A numerical sequence is an indexed list of numbers. It has a first term, a second one, etc. For instance, $u(0)=8, u(1)=6, u(2)=9$. >[!info] Formal Definition >A numerical sequence is a special type of function that maps every number $n$, such that $n \in \mathbb{N}$, to a number referred to as $u(n)$ or, more commonly, $u_n$, such that $u_n \in \mathbb{R}$. The $n$ **index** is sometimes called the **rank**. $u_n$ is interchangeably referred to as the sequence's term of index or rank $n$. The domain $\mathbb{N}$ usually starts with $n=0$, but sometimes it's more convenient to start with $n=1$. A numerical sequence can be generated either: - with an **explicit formula**, which defines $u_n$ in terms of its index $n$. e.g., $u(n)=-n^2+n+2$ - with a **recurrence relation**, where a term is defined by one or more preceding terms. The terms are computed iteratively (or step by step). The equation that defines this relationship, such as $u_{n+1}=\frac{u_n+3}{2}$, is called the recurrence relation. A recurrence relation must always be accompanied by an **initial condition** (or first term), e.g. $u_0=3$. ## Monotonicity A sequence is said to be monotonic if its terms consistently either decrease or increase. >[!info] Theorem >A sequence is said to be **monotonic**, if, $\forall n \in \mathbb{N}$, the sign of $u_{n+1}-u_n$ is always the same. If the sign of $u_{n+1}-u_n$ depends on the value of $n$, the sequence is **not monotonic**. - A monotonic sequence $u_n$ is increasing if $u_{n+1}\geq u_n$. - A monotonic sequence $u_n$ is strictly increasing if $u_{n+1}\gt u_n$. - A monotonic sequence $u_n$ is decreasing if $u_{n+1}\leq u_n$. - A monotonic sequence $u_n$ is strictly decreasing if $u_{n+1}\lt u_n$. ### Sign Test To understand the monotonicity (or behavior) of a sequence $u_n$ we analyze the sign of $u_{n+1}-u_n$. If $u_{n+1}-u_n \lt 0$, the sequence is strictly decreasing. >[!example] >**Prove that the sequence $u_n=\frac{3}{n+2}$ is increasing or decreasing.** > >The domain of the sequence is the set of natural numbers $\mathbb{N}=\{0,1,2,...\}$. > >To determine if the sequence is monotonic, we analyze the sign of $u_{n+1}-u_{n}$. > >$ >\begin{align} >u_{n+1}-u_{n} &=\frac{3}{n+3}-\frac{3}{n+2} >\\ >&=\frac{3(n+2)-3(n+3)}{(n+2)(n+3)} >\\ >&=\frac{3n+6-3n-9}{(n+2)(n+3)} >\\ >&=\frac{-3}{(n+2)(n+3)} >\end{align} >$ > >The numerator of $u_{n+1}-u_{n}$, $-3$, is negative. > >$\forall n \in \mathbb{N}$, $n+2 \gt 0$ and $n+3 \gt 0$. > >Therefore the denominator cannot be $0$ and is always positive. > >Therefore $u_{n+1}-u_{n}$ is a negative number divided by a positive number. > >Therefore, $\forall n \in \mathbb{N}$, $u_{n+1}-u_{n}\lt 0$. > >Therefore the sequence $u_n=\frac{3}{n+2}$ is strictly decreasing. >[!example] >**Prove that the sequence $v_n=\frac{3^n}{4^{n+2}}$ is increasing or decreasing.** > >The domain of the sequence is the set of natural numbers $\mathbb{N}=\{0,1,2,...\}$. > >To determine if the sequence is monotonic, we analyze the sign of $v_{n+1}-v_n$. > >$ >\begin{align} >v_{n+1}-v_n &= \frac{3^{n+1}}{4^{n+3}} - \frac{3^{n}}{4^{n+2}} >\\ >&= \frac{3^{n+1}}{4^{n+3}} - \frac{3^n \times 4}{4^{n+2} \times 4} >\\ >&=\frac{3^{n+1}-3^n \times 4}{4^{n+3}} >\\ >&=\frac{3^n \times 3 - 3^n \times 4}{4^{n+3}} >\\ >&=\frac{3^n \times (3 - 4)}{4^{n+3}} >\\ >&=-1 \cdot \frac{3^n}{4^{n+3}} >\\ >&=-\frac{3^n}{4^{n+3}} >\end{align} >$ > >$\forall n \in \mathbb{N}$, $3^n \gt 0$ and $4^{n+3} \gt 0$. > >Therefore, the fraction $\frac{3^n}{4^{n+3}}$ is a positive number divided by a positive number, which is always positive. > >Therefore the expression $-\frac{3^n}{4^{n+3}}$ is always negative. > >Therefore, $\forall n \in \mathbb{N}$, $v_{n+1}-v_n <0$. > >Therefore the sequence $v_n=\frac{3^n}{4^{n+2}}$ is strictly decreasing. ### Algebraic Identity We can simplify this method using the algebraic identity below: >[!info] Converting a Difference into a Product >$\forall b \neq 0$, $a - b = b \left( \frac{a}{b} - 1 \right)$ Applied to the sign test formula, we get: $v_{n+1}-v_{n}=v_{n} \left( \frac{v_{n+1}}{v_{n}} -1 \right)$ >[!example] >Going back to the sequence $v_n=\frac{3^n}{4^{n+2}}$, we see we can solve it by factoring out $v_n$. >$ >\begin{align} >v_{n} \left( \frac{v_{n+1}}{v_{n}} -1 \right) &= \frac{3^n}{4^{n+2}} \left( \frac{\frac{3^{n+1}}{4^{n+3}}}{\frac{3^n}{4^{n+2}}} - 1 \right) >\\ >&= \frac{3^n}{4^{n+2}} \left( \left( \frac{3^{n+1}}{4^{n+3}} \times \frac{4^{n+2}}{3^n}\right) -1 \right) >\\ >&= \frac{3^n}{4^{n+2}}\left( \left( \frac{3}{4} \right) -1 \right) >\\ >&= \frac{3^n}{4^{n+2}} \left( -\frac{1}{4} \right) >\end{align} >$ >The first factor, $\frac{3^n}{4^{n+2}}$ is the ratio of two positive numbers and is therefore always positive $\forall n \in \mathbb{N}$. > >The second factor is $-\frac{1}{4}$, a negative constant. > >The product of a positive number and a negative number is always negative. > >Therefore, $\forall n \in \mathbb{N}$, $v_{n+1}-v_n \lt 0$. > >Therefore, the sequence $v_n=\frac{3^n}{4^{n+2}}$ is strictly decreasing. By using this algebraic identity, we've made our analysis of the monotonicity of a sequence simpler. We can take this one step further. ### Ratio Test Let's go back to the algebraic identity. $ v_{n+1}-v_{n}=v_{n} \left( \frac{v_{n+1}}{v_{n}} -1 \right) $ To determine wether the sequence is increasing or decreasing, we need to know the sign of the left side. If all values of $v_n$ are positive, then $\frac{v_{n+1}}{v_{n}}$ will always be positive, too. This means we can simplify the problem entirely to comparing $\frac{v_{n+1}}{v_{n}}$ to $1$. If $\frac{v_{n+1}}{v_{n}} \gt 1$, $\left(\frac{v_{n+1}}{v_{n}} -1\right)$ yields a positive number, meaning the sequence is strictly increasing. If $\frac{v_{n+1}}{v_{n}} \lt 1$, $\left(\frac{v_{n+1}}{v_{n}} -1\right)$ yields a negative number, meaning the sequence is strictly decreasing. If $\frac{v_{n+1}}{v_{n}} = 1$, the sequence is constant, because $\left(\frac{v_{n+1}}{v_{n}} -1\right)$ yields $0$. >[!info] Theorem >In other words, where we're able to determine that $\forall n \in \mathbb{N}$, $v_n \gt 0$, we can reduce the analysis of the monotonicity of the sequence to comparing $\left(\frac{v_{n+1}}{v_{n}}\right)$ to $1$. This method turns out to be very efficient where applicable. >[!example] >Going back once more to the sequence $v_n=\frac{3^n}{4^{n+2}}$, we prove its monotonicity by factoring out $v_n$. >$ >\begin{align} >\frac{v_{n+1}}{v_n} &= \frac{3^{n+1}}{4^{n+3}} \times \frac{4^{n+2}}{3^n} >\\ >&= \frac{3 \cdot 3^n}{4 \cdot 4^{n+2}} \times \frac{4^{n+2}}{3^n} >\\ >&=\frac{3}{4} >\end{align} >$ > >Since $\frac{v_{n+1}}{v_n} = \frac{3}{4} < 1$, the sequence $v_n$ is **strictly decreasing**. ## Arithmetic Sequences A numerical sequence is said to be arithmetic if, $\forall n$, $u_{n+1}-u_n$ is a constant value. $ u_{n+1}-u_n=d $ This constant is called the **common difference** of the sequence, usually denoted by $d$. ## Geometric Sequences A numerical sequence is said to be geometric if the ratio of any term to its preceding term is a constant value. In other words, you always obtain the next term by multiplying by the same real number. $ \frac{u_{n+1}}{n}=r $ This number is called the **common ratio** of the sequence, usually denoted by $r$.