Sequences and Their Limits

Consider the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \{ \frac{1}{n} \right \}_{n=1}^{\infty} = \left \{ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ... \right \}} . As Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \to \infty} , it appears as though ${\frac {1}{n}}\to 0$. In fact, we know that this is true since $\lim _{n\to \infty }{\frac {1}{n}}=0$. We will now formalize the definition of a limit with regards to sequences.

Definition: If $\{a_{n}\}$ is a sequence, then $\lim _{n\to \infty }a_{n}=L$ means that for every $\epsilon >0$ there exists a corresponding $N\in \mathbb {N}$ such that if $n\geq N$, then $\mid a_{n}-L\mid <\epsilon$. If this limit exists, then we say that the sequence $\{a_{n}\}$ Converges, and if this limit doesn't exist then we say say the sequence $\{a_{n}\}$ Diverges.

We note that our definition of the limit of a sequence is very similar to the limit of a function, in fact, we can think of a sequence as a function whose domain is the set of natural numbers $\mathbb {N}$. From this notion, we obtain the very important theorem:

Theorem 1: If $\{a_{n}\}$ is a sequence and a function $f(x)$, then if $f(n)=a_{n}$ where $n\in \mathbb {N}$ and $\lim _{x\to \infty }f(x)=L$, then $\lim _{n\to \infty }f(n)=L$.

• Proof of Theorem 1: Let $\epsilon >0$ be given. We know that $\lim _{x\to \infty }f(x)=L$ which implies that $\forall \epsilon >0\exists N\in \mathbb {R}$ such that if $x\geq k$ then $\mid f(x)-L\mid <\epsilon$.

• Now we want to show that $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then $\mid a_{n}-L\mid <\epsilon$. We will choose $N=\mathrm {max} \{1,\lceil k\rceil \}$. This ensures that $N$ is an integer.

• Now since $N\geq k$ then it follows that if $\forall n\geq N$ then $\mid f(n)-L\mid <\epsilon$. But $f(n)=a_{n}$ and so $\mid a_{n}-L\mid <\epsilon$ so $\lim _{n\to \infty }a_{n}=L$. $\blacksquare$

Important Note: The converse of this theorem is not implied to be true! That is if $f(n)=a_{n}$ and $\lim _{n\to \infty }f(n)=L$, then this does NOT imply that $\lim _{x\to \infty }f(x)=L$. For example, consider the sequence $\{\sin(2\pi n)\}=\{0,0,0,...\}$. Clearly this sequence converges at 0. However, the function $f(x)=\sin(2\pi x)$ does not converge, instead, it diverges as it oscillates between $-1$ and $1$.

For example, consider the sequence $\left\{{\frac {n+2}{n}}\right\}_{n=1}^{\infty }$. If we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(n) = \frac{n + 2}{n}} be a function whose domain is the natural numbers, then we calculate the limit of this function like we have in the past, namely:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{n + 2}{n} = \lim_{n \to \infty} 1 + \frac{2}{n} = 1 \end{align}}

Therefore the limit of our sequence $f(n)=a_{n}$ is 1, that is, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \{ \frac{n + 2}{n} \right \}_{n=1}^{\infty}} converges to 1 as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \to \infty} .

Now let's look at another major theorem.

Theorem 2: If $\lim _{n\to \infty }a_{n}=L$ and a function $f$ is continuous at $L$, then $\lim _{n\to \infty }f(a_{n})=f\left(\lim _{n\to \infty }a_{n}\right)=f(L)$.

• Proof of Theorem 2: If $f$ is a continuous function at $x=L$, then we know that $f(L)$ is defined and that $\lim _{x\to L}f(x)=f(L)$. By the definition of a limit, $\forall \epsilon >0\exists \delta >0$ such that if $0<\mid x-L\mid <\delta$ then $\mid f(x)-f(L)\mid <\epsilon$. Let $x=a_{n}$ so then if $0<\mid a_{n}-L\mid <\delta$ then $\mid f(a_{n})-f(L)\mid <\epsilon$.

• We want to show that $\lim _{n\to \infty }f(a_{n})=f(L)$, that is $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then $\mid f(a_{n})-f(L)\mid <\epsilon$, which is what we showed above.

We will now look at some important limit laws regarding sequences

Limit Laws of Convergent Sequences

We will now look at some very important limit laws regarding convergent sequences, all of which are analogous to the limit laws for functions that we already know of.

• Law 1 (Addition Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} b_n = B} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = A + B} .

• Law 2 (Difference Law of Convergent Sequences): If the limits of the sequences Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ a_n \}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ b_n \}} are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} b_n = B} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} (a_n - b_n) = \lim_{n \to \infty} a_n - \lim_{n \to \infty} b_n = A - B} .

• Law 3 (Product Law of Convergent Sequences): If the limits of the sequences Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ a_n \}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ b_n \}} are convergent, that is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} a_n = A} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} b_n = B} , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} (a_nb_n) = \lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = AB} .

• Law 4 (Quotient Law of Convergent Sequences): If the limits of the sequences Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ a_n \}} and $\{b_{n}\}$ are convergent, that is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} a_n = A} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} b_n = B} , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{\lim_{n \to \infty} a_n}{\lim_{n \to \infty} b_n} = \frac{A}{B}} provided that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} b_n \neq 0} .

• Law 5 (Constant Multiple Law of Convergent Sequences): If the limit of the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ a_n \}} is convergent, that is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} a_n = A} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} is a constant, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} ka_n = k \lim_{n \to \infty} a_n = kA} .

• Law 6 (Power Law of Convergent Sequences): If the limit of the sequence $\{a_{n}\}$ is convergent, that is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} a_n = A} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} is a non-negative integer, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} (a_n)^k = \left ( \lim_{n \to \infty} a_n \right )^k = (A)^k} provided that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in \mathbb{N}} .

• Law 7 (Squeeze Theorem for Convergent Sequences): If the limits of the sequences Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ a_n \}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ b_n \}} , and $\{c_{n}\}$ are convergent and $a_{n}\leq b_{n}\leq c_{n}$ is true always after some $n^{\mathrm {th} }$ term, if $\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }c_{n}=L$, then $\lim _{n\to \infty }b_{n}=L$.

Example 1

Determine whether the sequence $\left\{{\frac {x^{2}-1}{x+1}}\right\}_{n=1}^{\infty }$ is convergent or divergent.

Let $f(n)={\frac {n^{2}-1}{n+1}}$ be a function analogous to our sequence. When we factor the numerator and cancel like-terms, we get that $\lim _{n\to \infty }f(n)=\lim _{n\to \infty }{\frac {n^{2}-1}{n+1}}=\lim _{n\to \infty }{\frac {(n+1)(n-1)}{n+1}}=\lim _{n\to \infty }n-1=\infty$. Therefore the sequence $\left\{{\frac {x^{2}-1}{x+1}}\right\}_{n=1}^{\infty }$ is divergent.